Polynomials And Factorisation Class 9th Mathematics AP Board Solution

Class 9th Mathematics AP Board Solution
Exercise 2.1
  1. Find the degree of each of the polynomials given below (i) x^5 - x^4 + 3 (ii)…
  2. Which of the following expressions are polynomials in one variable and which are…
  3. Write the coefficient of x^3 in each of the following (i) x^3 + x + 1 (ii) 2 -…
  4. Classify the following as linear, quadratic and cubic polynomials (i) 5x^2 + x -…
  5. Write whether the following statements are True or False. Justify your answer…
  6. Give one example each of a monomial and trinomial of degree 10.
Exercise 2.2
  1. Find the value of the polynomial 4x^2 - 5x + 3, when (i) x = 0 (ii) x = -1 (iii)…
  2. p(x) = x^2 - x +1 Find p(0), p(1) and p(2) for each of the following…
  3. p(y) = 2 + y + 2y^2 - y^3 Find p(0), p(1) and p(2) for each of the following…
  4. p(z) = z^3 Find p(0), p(1) and p(2) for each of the following polynomials.…
  5. p(t) = (t - 1) (t + 1) Find p(0), p(1) and p(2) for each of the following…
  6. p(x) = x^2 - 3x + 2 Find p(0), p(1) and p(2) for each of the following…
  7. p(x) = 2x + 1; x = - 1/2 Verify whether the values of x given in each case are…
  8. p(x) = 5x - π; x = -3/2 Verify whether the values of x given in each case are…
  9. p(x) = x^2 - 1; x = ±1 Verify whether the values of x given in each case are…
  10. p(x) = (x - 1)(x + 2); x = -1, -2 Verify whether the values of x given in each…
  11. p(y) = y^2 ; y = 0 Verify whether the values of x given in each case are the…
  12. p(x) = ax + b ; x = - b/a Verify whether the values of x given in each case are…
  13. f(x) = 3x^2 - 1; x = - 1/root 3 , 2/root 3 Verify whether the values of x given…
  14. f (x) = 2x - 1, x = 1/2 , -1/2 Verify whether the values of x given in each…
  15. f(x) = x + 2 Find the zero of the polynomial in each of the following cases.…
  16. f(x) = x - 2 Find the zero of the polynomial in each of the following cases.…
  17. f(x) = 2x + 3 Find the zero of the polynomial in each of the following cases.…
  18. f(x) = 2x - 3 Find the zero of the polynomial in each of the following cases.…
  19. f(x) = x^2 Find the zero of the polynomial in each of the following cases.…
  20. f(x) = px, p ≠ 0 Find the zero of the polynomial in each of the following…
  21. f(x) = px + q, p ≠ 0, p q are real numbers. Find the zero of the polynomial in…
  22. If 2 is a zero of the polynomial p(x) = 2x^2 - 3x + 7a, find the value of a.…
  23. If 0 and 1 are the zeroes of the polynomial f(x) = 2x^3 - 3x^2 + ax + b, find…
Exercise 2.3
  1. x + 1 Find the remainder when x^3 + 3x^2 + 3x + 1 is divided by the following…
  2. x - 1/2 Find the remainder when x^3 + 3x^2 + 3x + 1 is divided by the following…
  3. x Find the remainder when x^3 + 3x^2 + 3x + 1 is divided by the following…
  4. x + π Find the remainder when x^3 + 3x^2 + 3x + 1 is divided by the following…
  5. 5 + 2x Find the remainder when x^3 + 3x^2 + 3x + 1 is divided by the following…
  6. Find the remainder when x^3 - px^2 + 6x - p is divided by x - p.
  7. Find the remainder when 2x^2 - 3x + 5 is divided by 2x - 3. Does it exactly…
  8. Find the remainder when 9x^3 - 3x^2 + x - 5 is divided by x - 2/3…
  9. If the polynomials 2x^3 + ax^2 + 3x - 5 and x^3 + x^2 - 4x + a leave the same…
  10. If the polynomials x^3 + ax^2 + 5 and x^3 - 2x^2 + a are divided by (x + 2)…
  11. Find the remainder when f (x) = x^4 - 3x^2 + 4 is divided by g(x)= x - 2 and…
  12. Find the remainder when p(x) = x^3 - 6x^2 + 14x - 3 is divided by g(x) = 1 - 2x…
  13. When a polynomial 2x^3 +3x^2 + ax + b is divided by (x - 2) leaves remainder 2,…
Exercise 2.4
  1. x^3 - x^2 - x + 1 Determine which of the following polynomials has (x + 1) as a…
  2. x^4 - x^3 + x^2 - x + 1 Determine which of the following polynomials has (x +…
  3. x^4 + 2x^3 + 2x^2 + x + 1 Determine which of the following polynomials has (x +…
  4. x^3 - x^2 -(3 - √3) x + √3 Determine which of the following polynomials has (x…
  5. f(x) = 5x^3 + x^2 - 5x - 1, g(x) = x + 1 Use the Factor Theorem to determine…
  6. f(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 1 Use the Factor Theorem to determine…
  7. f(x) = x^3 - 4x^2 + x + 6, g(x) = x - 2 Use the Factor Theorem to determine…
  8. f(x) = 3x^3 + x^2 - 20x + 12, g(x) = 3x-2 Use the Factor Theorem to determine…
  9. f(x) = 4x^3 + 20x^2 + 33x + 18, g(x) = 2x + 3 Use the Factor Theorem to…
  10. Show that (x - 2), (x + 3) and (x - 4) are factors of x^3 - 3x^2 - 10x + 24.…
  11. Show that (x + 4), (x - 3) and (x - 7) are factors of x^3 - 6x^2 - 19x + 84.…
  12. If both (x - 2) and (x - 1/2) are factors of px^2 + 5x + r, show that p = r.…
  13. If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b…
  14. x^3 - 2x^2 - x + 2 Factorize
  15. x^3 - 3x^2 - 9x - 5 Factorize
  16. x^3 + 13x^2 + 32x + 20 Factorize
  17. y^3 + y^2 - y - 1 Factorize
  18. If ax^2 + bx + c and bx^2 + ax + c have a common factor x + 1 then show that c =…
  19. If x^2 - x - 6 and x^2 + 3x - 18 have a common factor (x - a) then find the…
  20. If (y - 3) is a factor of y^3 - 2y^2 - 9y + 18 then find the other two factors.…
Exercise 2.5
  1. (x + 5) (x + 2) Use suitable identities to find the following products…
  2. (x - 5) (x - 5) Use suitable identities to find the following products…
  3. (3x + 2)(3x - 2) Use suitable identities to find the following products…
  4. (x^2 + 1/x^2) (x^2 - 1/x^2) Use suitable identities to find the following…
  5. (1 + x) (1 + x) Use suitable identities to find the following products…
  6. 101 × 99 Evaluate the following products without actual multiplication.…
  7. 999 × 999 Evaluate the following products without actual multiplication.…
  8. 50 1/2 x 49 1/2 Evaluate the following products without actual multiplication.…
  9. 501 × 501 Evaluate the following products without actual multiplication.…
  10. 30.5 × 29.5 Evaluate the following products without actual multiplication.…
  11. 16x^2 + 24xy + 9y^2 Factorise the following using appropriate identities.…
  12. 4y^2 - 4y + 1 Factorise the following using appropriate identities.…
  13. 4x^2 - y^2/25 Factorise the following using appropriate identities.…
  14. 18a^2 - 50 Factorise the following using appropriate identities.
  15. x^2 + 5x + 6 Factorise the following using appropriate identities.…
  16. 3p^2 - 24p + 36 Factorise the following using appropriate identities.…
  17. (x + 2y + 4z)^2 Expand each of the following, using suitable identities…
  18. (2a - 3b)^3 Expand each of the following, using suitable identities…
  19. (-2a + 5b - 3c)^2 Expand each of the following, using suitable identities…
  20. (a/4 - b/2 + 1)^2 Expand each of the following, using suitable identities…
  21. (p + 1)^3 Expand each of the following, using suitable identities…
  22. (x - 2/3 y)^3 Expand each of the following, using suitable identities…
  23. 25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz Factorise
  24. 9a^2 + 4b^2 + 16c^2 + 12ab - 16bc - 24ca Factorise

Exercise 2.1
Question 1.

Find the degree of each of the polynomials given below

(i) x5 - x4 + 3

(ii) x2 + x - 5

(iii) 5

(iv) 3x6 + 6y3 - 7

(v) 4 - y2

(vi) 5t - √3


Answer:

Degree of p(x) is the highest power of x in p(x).


(i) The highest power of x in x5 - x4 + 3 is 5.


∴ The degree of x5 - x4 + 3 is 5.


(ii) The highest power of x in x2 + x - 5 is 2.


∴ The degree of x2 + x - 5 is 2.


(iii) The highest power of x in 5 is 0(∵ there is no term of x).


∴ The degree of 5 is 0.


(iv) The highest power of x in 3x6 + 6y3 - 7 is 6.


∴ The degree of 3x6 + 6y3 - 7 is 6.


(v) The highest power of y in 4 - y2 is 2.


∴ The degree of 4 - y2 is 2.


(vi) The highest power of t in 5t – √3 is 1.


∴ The degree of 5t – √3 is 1.



Question 2.

Which of the following expressions are polynomials in one variable and which are not? Give reasons for your answer.

(i) 3x2 - 2x + 5

(ii) x2 + √2

(iii) p2 - 3p + q

(iv) 

(v) 

(vi) x100 + y100


Answer:

(i) 3x2 - 2x + 5 has only one variable that is x.


∴ yes, it is a polynomial in one variable.


(ii) x2 + √2 has only one variable that is x.


∴ yes, it is a polynomial in one variable.


(iii) p2 – 3p + q has two variables that are p and q.


∴ no, it is not a polynomial in one variable.


(iv)  has a negative exponent of y.


∴ no, it is not a polynomial.


(v) The exponent of x in 5√x + x√5 is 1/2 which is not a non-negative integer


∴ no, it is not a polynomial.


(vi) x100 + y100 has two variables that are x and y.


∴ no, it is not a polynomial in one variable.



Question 3.

Write the coefficient of x3 in each of the following

(i) x3 + x + 1 (ii) 2 - x3 + x2

(iii)  (iv) 2x3 + 5

(v)  (vi) 

(vii) 2x2 + 5 (vi) 4


Answer:

A coefficient is a multiplicative factor in some term of a polynomial. It is the constant written before the variable.


Therefore,


(i) The constant written before x3 in x3 + x + 1 is 1.


∴ The coefficient of x3 in x3 + x + 1 is 1.


(ii) The constant written before x3 in 2 – x3 + x2 is -1.


∴ The coefficient of x3 in 2 – x3 + x2 is -1.


(iii) The constant written before x3 in √2x3 + 5 is √2.


∴ The coefficient of x3 in √2x3 + 5 is √2.


(iv) The constant written before x3 in 2x3 + 5 is 2.


∴ The coefficient of x3 in 2x3 + 5 is 2.


(v) The constant written before x3 in  is.


∴ The coefficient of x3 in  is.


(vi) The constant written before x3 in  is.


∴ The coefficient of x3 in  is.


(vii) The term x3 does not exist in 2x2 + 5.


∴ The coefficient of x3 in 2x2 + 5 is 0.


(viii) The term x3 does not exist in 4.


∴ The coefficient of x3 in 4 is 0.



Question 4.

Classify the following as linear, quadratic and cubic polynomials

(i) 5x2 + x - 7 (ii) x - x3

(iii) x2 + x + 4 (iv) x - 1

(v) 3p (vi) πr2


Answer:

(i) A quadratic polynomial is a polynomial of degree 2


∵ the degree of 5x2 + x – 7 is 2


∴ 5x2 + x – 7 is a quadratic polynomial.


(ii) A cubic polynomial is a polynomial of degree 3


∵ the degree of 5x2 + x – 7 is 3


∴ 5x2 + x – 7 is a cubic polynomial.


(iii) A quadratic polynomial is a polynomial of degree 2


∵ the degree of x2 + x + 4 is 2


∴ x2 + x + 4 is a quadratic polynomial.


(iv) A linear polynomial is a polynomial of degree 1


∵ the degree of x – 1 is 1


∴ x – 1 is a linear polynomial.


(v) A linear polynomial is a polynomial of degree 1


∵ the degree of 3p is 1


∴ 3p is a linear polynomial.


(vi) A quadratic polynomial is a polynomial of degree 2


∵ the degree of πr2 is 2


∴ πr2 is a quadratic polynomial.



Question 5.

Write whether the following statements are True or False. Justify your answer

(i) A binomial can have at the most two terms

(ii) Every polynomial is a binomial

(iii) A binomial may have degree 3

(iv) Degree of zero polynomial is zero

(v) The degree of x2 + 2xy + y2 is 2

(vi) πr2 is monomial.


Answer:

(i) A polynomial with two terms is called a binomial.


∴ The statement is true.


(ii) A polynomial can have more than two terms.


∴ The statement is false.


(iii) A binomial should have two terms, the degree of those terms can be any integer.


∴ The statement is true.


(iv) The constant polynomial whose coefficients are all equal to 0, is called a zero polynomial. Its degree can be any integer.


∴ The statement is false.


(v) The highest power in x2 + 2xy + y2 is 2, therefore its degree is 2.


∴ The statement is true.


(vi) A monomial is a polynomial which has only one term.


∵ πr2 has only one term


∴ The statement is true.



Question 6.

Give one example each of a monomial and trinomial of degree 10.


Answer:

A monomial is a polynomial which has only one term, and the degree is the highest power of the variable. Therefore, an example of a monomial of degree 10 is 3x10.


A trinomial is a polynomial which has three terms, and the degree is the highest power of the variable. Therefore, example of a trinomial of degree 10 is 3x10 + 2x2 + 5.




Exercise 2.2
Question 1.

Find the value of the polynomial 4x2 - 5x + 3, when

(i) x = 0 (ii) x = -1

(iii) x = 2 (iv) 


Answer:

(i) p(x) = 4x2 – 5x + 3


⇒ p(0) = 4(0)2 – 5(0) + 3


⇒ p(0) = 0– 0 + 3


⇒ p(0) = 3


(ii) p(x) = 4x2 – 5x + 3


⇒ p(-1) = 4(-1)2 – 5(-1) + 3


⇒ p(-1) = 4 × 1– (-5) + 3


⇒ p(-1) = 4 +5 + 3


⇒ p(-1) = 12


(iii) p(x) = 4x2 – 5x + 3


⇒ p(2) = 4(2)2 – 5(2) + 3


⇒ p(2) = 4 × 4– 10 + 3


⇒ p(2) = 16– 10 + 3


⇒ p(2) = 9


(iv) p(x) = 4x2 – 5x + 3








Question 2.

Find p(0), p(1) and p(2) for each of the following polynomials.

p(x) = x2 - x +1


Answer:

p(x) = x2 – x + 1


⇒ p(0) = (0)2 – 0 + 1


⇒ p(0) = 1


And,


⇒ p(1) = (1)2 – 1 + 1


⇒ p(1) = 1– 1 + 1


⇒ p(1) = 1


And,


⇒ p(2) = (2)2 – 2 + 1


⇒ p(2) = 4– 2 + 1


⇒ p(2) = 3



Question 3.

Find p(0), p(1) and p(2) for each of the following polynomials.

p(y) = 2 + y + 2y2 - y3


Answer:

p(y) = 2 + y + 2y2 – y3


⇒ p(0) = 2 + 0 + 2(0)2 – (0)3


⇒ p(0) = 2 + 0 + 0– 0


⇒ p(0) = 2


And,


⇒ p(1) = 2 + 1 + 2(1)2 – (1)3


⇒ p(1) = 2 + 1 + 2– 1


⇒ p(1) = 4


And,


⇒ p(2) = 2 + 2 + 2(2)2 – (2)3


⇒ p(2) = 2 + 2 + 8– 8


⇒ p(2) = 4



Question 4.

Find p(0), p(1) and p(2) for each of the following polynomials.

p(z) = z3


Answer:

p(z) = z3


⇒ p(0) = 03


⇒ p(0) = 0


And,


⇒ p(1) = 13


⇒ p(1) = 1


And,


⇒ p(2) = 23


⇒ p(2) = 8



Question 5.

Find p(0), p(1) and p(2) for each of the following polynomials.
p(t) = (t - 1) (t + 1)


Answer:

p(t) = (t - 1) (t + 1)


p(t) = t2 + t - t - 1

⇒ p(t) = t2 – 1


⇒ p(0) = (0)2 – 1


⇒ p(0) = 0– 1


⇒ p(0) = -1


And,


⇒ p(1) = (1)2 – 1


⇒ p(1) = 1– 1


⇒ p(1) = 0


And,


⇒ p(2) = (2)2 – 1


⇒ p(2) = 4– 1


⇒ p(2) = 3


Question 6.

Find p(0), p(1) and p(2) for each of the following polynomials.

p(x) = x2 - 3x + 2


Answer:

p(t) = x2 - 3x + 2


⇒ p(0) = (0)2 – 3(0) + 2


⇒ p(0) = 0– 0 + 2


⇒ p(0) = 2


And,


⇒ p(1) = (1)2 – 3(1) + 2


⇒ p(1) = 1– 3 + 2


⇒ p(0) = 0


And,


⇒ p(2) = (2)2 – 3(2) + 2


⇒ p(2) = 4– 6 + 2


⇒ p(2) = 0



Question 7.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = 2x + 1; 


Answer:

p(x) = 2x + 1


⇒ p(-1/2) = 2(-1/2) + 1


⇒ p(-1/2) = -1 + 1


⇒ p(-1/2) = 0


∴ Yes x = -1/2 is the zero of polynomial 2x + 1.



Question 8.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = 5x - π; 


Answer:

p(x) = 5x - π





∴ No x = - is not the zero of polynomial 5x – π.



Question 9.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = x2 - 1; x = ±1


Answer:

p(x) = x2 - 1


⇒ p(-1) = (-1)2 – 1


⇒ p(-1) = 1 – 1


⇒ p(-1) = 0


∴ Yes x = -1 is the zero of polynomial x2 – 1.


And,


⇒ p(1) = (1)2 – 1


⇒ p(1) = 1 – 1


⇒ p(1) = 0


∴ Yes x = 1 is the zero of polynomial x2 – 1.



Question 10.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = (x - 1)(x + 2); x = -1, -2


Answer:

p(x) = (x - 1)(x + 2)


⇒ p(-1) = (-1 - 1)(-1 + 2);


⇒ p(-1) = -2 × 1


⇒ p(-1) = -2


∴ No x = -1 is not the zero of polynomial (x - 1)(x + 2).


And,


⇒ p(-2) = (-2 - 1)(-2 + 2);


⇒ p(-2) = -3 × 0


⇒ p(-2) = 0


∴ Yes x = -2 is not the zero of polynomial (x - 1)(x + 2).



Question 11.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(y) = y2; y = 0


Answer:

p(y) = y2


⇒ p(0) = 02


⇒ p(0) = 0


∴ Yes y = 0 is the zero of polynomial y2.



Question 12.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = ax + b ; 


Answer:

p(x) = ax + b





∴ Yes  is the zero of polynomial ax + b.



Question 13.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

f(x) = 3x2 - 1; x = - 


Answer:

f(x) = 3x2 - 1






∴ Yes  is the zero of polynomial 3x2 - 1.






∴ No  is not the zero of polynomial 3x2 - 1.



Question 14.

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

f (x) = 2x - 1, x = 


Answer:

f(x) = 2x - 1






∴ Yes  is the zero of polynomial 2x - 1.






∴ No  is not the zero of polynomial 2x - 1.



Question 15.

Find the zero of the polynomial in each of the following cases.

f(x) = x + 2


Answer:

f(x) = x + 2


f(x) = 0


⇒ x + 2 = 0


⇒ x = 0 – 2


⇒ x = -2


∴ x = -2 is the zero of the polynomial x + 2.



Question 16.

Find the zero of the polynomial in each of the following cases.

f(x) = x - 2


Answer:

f(x) = x - 2


f(x) = 0


⇒ x – 2 = 0


⇒ x = 0 + 2


⇒ x = 2


∴ x = 2 is the zero of the polynomial x – 2.



Question 17.

Find the zero of the polynomial in each of the following cases.

f(x) = 2x + 3


Answer:

f(x) = 2x + 3


f(x) = 0


⇒ 2x + 3 = 0


⇒ 2x = 0 – 3


⇒ 2x = -3



∴  is the zero of the polynomial 2x + 3.



Question 18.

Find the zero of the polynomial in each of the following cases.

f(x) = 2x - 3


Answer:

f(x) = 2x – 3


f(x) = 0


⇒ 2x – 3 = 0


⇒ 2x = 0 + 3


⇒ 2x = 3



∴  is the zero of the polynomial 2x – 3.



Question 19.

Find the zero of the polynomial in each of the following cases.

f(x) = x2


Answer:

f(x) = x2


f(x) = 0


⇒ x2 = 0


⇒ x = 0


∴ x = 0 is the zero of the polynomial x2.



Question 20.

Find the zero of the polynomial in each of the following cases.

f(x) = px, p ≠ 0


Answer:

f(x) = px, p ≠ 0


f(x) = 0


⇒ px = 0


⇒ x = 0


∴ x = 0 is the zero of the polynomial px.



Question 21.

Find the zero of the polynomial in each of the following cases.

f(x) = px + q, p ≠ 0, p q are real numbers.


Answer:

f(x) = px + q, p ≠ 0, p q are real numbers.


f(x) = 0


⇒ px + q = 0


⇒ px = -q



∴  is the zero of the polynomial px + q.



Question 22.

If 2 is a zero of the polynomial p(x) = 2x2 - 3x + 7a, find the value of a.


Answer:

∵ 2 is the zeroes of the polynomial p(x) = 2x2 - 3x + 7a


∴ p(2) = 0


Now,


p(x) = 2x2 - 3x + 7a


⇒p(2) = 2(2)2 – 3(2)+ 7a


⇒ 2 × 4– 3 × 2 + 7a = 0


⇒ 8– 6 + 7a = 0


⇒ 2 + 7a = 0


⇒ 7a = -2




Question 23.

If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 - 3x2 + ax + b, find the values of a and b.


Answer:

∵ 0 and 1 are the zeroes of the polynomial f(x) = 2x3-3x2+ ax + b


∴ f(0) = 0 and f(1) = 1


Now,


f(x) = 2x3-3x2+ ax + b


⇒ f(0) = 2(0)3 – 3(0)2+ a(0) + 0


⇒ 2 × 0– 3 × 0 + a × 0 + b = 0


⇒ 0– 0 + 0 + b = 0


⇒ b = 0


And,


⇒ f(1) = 2(1)3 – 3(1)2+ a(1) +1


⇒ 2 × 1– 3 × 1 + a × 1 + b = 0


⇒ 2– 3 + a + b = 0


⇒ 2– 3 + a + 0 = 0 [∵ b = 0]


⇒ -1 + a = 0


⇒ a = 1




Exercise 2.3
Question 1.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x + 1


Answer:

Let p(x) = x3 + 3x2 + 3x + 1


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x+1 is p(–1)


p(–1) = (–1)3 + 3(–1)2 + 3(–1) +1


⇒ p(–1) = –1 + 3 – 3 + 1 = 0


∴ Remainder of x3 + 3x2 + 3x + 1 when divided by x+1 is 0



Question 2.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:



Answer:

Let p(x) = x3 + 3x2 + 3x + 1


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by  is p(1/2)




∴ Remainder of x3 + 3x2 + 3x + 1 when divided by  is 



Question 3.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x


Answer:

Let p(x) = x3 + 3x2 + 3x + 1


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x is p(0)


P(0) = (0)3+ 3(0)2+ 3(0) + 1


= 1


∴ Remainder of x3 + 3x2 + 3x + 1 when divided by x is 1



Question 4.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x + π


Answer:

Let p(x) = x3 + 3x2 + 3x + 1


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x + π is p(–π)


p(–π) = (–π)3+ 3(–π)2+ 3(–π) + 1


⇒ p(–π)= –π3 + 3π2 –3π + 1


∴ Remainder of x3 + 3x2 + 3x + 1 when divided by x+ π is –π3 + 3π2 –3π + 1



Question 5.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

5 + 2x


Answer:

Let p(x) = x3 + 3x2 + 3x + 1


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by 5 + 2x is 




∴ Remainder of x3 + 3x2 + 3x + 1 when divided by 5 + 2x is 



Question 6.

Find the remainder when x3 – px2 + 6x – p is divided by x – p.


Answer:

Let q(x) = x3 – px2 + 6x – p


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of q(x) when divided by x – p is q(p)


q(p) = (p)3– p(p)2+ 6(p) – p


⇒ q(p) = p3 – p3 + 6p – p


∴ Remainder of x3 – px2 + 6x – p when divided by x – p is 5p



Question 7.

Find the remainder when 2x2 – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial? State reason.


Answer:

Let p(x) = 2x2 – 3x + 5


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by 2x – 3 is 



= 5

⇒ Remainder of 2x2 – 3x + 5 when divided by 2x – 3 is 5.


As on dividing the given polynomial by 2x – 3, we get a non–zero remainder, therefore, 2x – 3 does not completely divide the polynomial.


∴ It is not a factor.


Question 8.

Find the remainder when 9x3 – 3x2 + x – 5 is divided by 


Answer:

Let p(x) = 9x3 – 3x2 + x – 5


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by  is 




⇒ Remainder of 9x3 – 3x2 + x – 5 when divided by  is –3



Question 9.

If the polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x + a leave the same remainder when divided by x – 2, find the value of a.


Answer:

Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 4x + a


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)


⇒ p(2) = 2(2)3 +a(2)2 + 3(2) – 5


⇒p(2) = 16 + 4a +6 – 5


⇒p(2) = 17 + 4a


Similarly, q(2) = (2)3 + (2)2 + –4(2) + a


⇒ q(2) = 8 + 4 –8 + a


⇒ q(2) = 4 + a


Since they both leave the same remainder, so p(2) = q(2)


⇒ 17 + 4a = 4 + a


⇒ 13 = 3a



∴ The value of a is –13/3



Question 10.

If the polynomials x3 + ax2 + 5 and x3 – 2x2 + a are divided by (x + 2) leave the same remainder, find the value of a.


Answer:

Let p(x) = x3 + ax2 + 5 and q(x) = x3 – 2x2 + a


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x + 2 is p(–2). Similarly, Remainder of q(x) when divided by x + 2 is q(–2)


⇒ p(–2) = (–2)3 +a(–2)2 + 5


⇒p(–2) = –8 + 4a + 5


⇒p(–2) = –3 + 4a


Similarly, q(–2) = (–2)3 – 2(–2)2 + a


⇒ q(–2) = –8 –8 + a


⇒ q(–2) = –16 + a


Since they both leave the same remainder, so p(–2) = q(–2)


⇒ –3 + 4a = –16 + a


⇒ –13 = 3a



∴ The value of a is –13/3



Question 11.

Find the remainder when f (x) = x4 – 3x2 + 4 is divided by g(x)= x – 2 and verify the result by actual division.


Answer:

As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


Therefore, remainder when f(x) is divided by g(x) is f(2)


f(2) = 24 – 3(2)2 + 4


⇒ f(2) = 16 – 12 + 4 = 8


∴ the remainder when x4 – 3x2 + 4 is divided by x – 2 is 8



Question 12.

Find the remainder when p(x) = x3 – 6x2 + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division.


Answer:

Given: p(x) = x3 – 6x2 + 14x – 3 and g(x) = 1 – 2x


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


Therefore, remainder when p(x) is divided by g(x) is 





∴ the remainder when x3 – 6x2 + 14x – 3 is divided by 1 – 2x is 
Result Verification:


We can see that the remainder is .

Hence, verified.


Question 13.

When a polynomial 2x3 +3x2 + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder –2. Find a and b.


Answer:

Let p(x) = 2x3 +3x2 + ax + b


As we know by Remainder Theorem,


If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)


⇒ Remainder of p(x) when divided by x – 2 is p(2)


p(2) = 2(2)3 +3(2)2 + a(2) + b


⇒ p(2) = 16 + 12 + 2a + b


Also, it is given that p(2) = 2, on substituting value above, wev get,


2 = 28 + 2a + b


⇒ 2a + b = –26 ---------- (A)


Similarly,


Remainder of p(x) when divided by x + 2 is p(–2)


p(–2) = 2(–2)3 +3(–2)2 + a(–2) + b


⇒ p(–2) = –16 + 12 – 2a + b


Also, it is given that p(–2) = –2, on substituting value above, we get,


–2 = –4 – 2a + b


⇒ – 2a + b = 2 ---------- (B)


On solving the above two equ. (A) and (b), we get,


a = –7 and b = –12


∴ Value of a and b is –7 and –12 respectively.




Exercise 2.4
Question 1.

Determine which of the following polynomials has (x + 1) as a factor.

x3 – x2 – x + 1


Answer:

Let f(x) = x3 – x2 – x + 1


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = (–1)3 – (–1)2 – (–1) + 1


⇒ f(–1) = –1 –1 +1 +1


⇒ f(–1) = 0


As, f(–1) is equal to zero, therefore (x+1) is a factor x3 – x2 – x + 1



Question 2.

Determine which of the following polynomials has (x + 1) as a factor.

x4 – x3 + x2 – x + 1


Answer:

Let f(x) = x4 – x3 + x2 – x + 1


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = (–1)4 – (–1)3 + (–1)2 – (–1) + 1


⇒ f(–1) = 1 + 1 + 1 + 1 + 1


⇒ f(–1) = 5


As, f(–1) is not equal to zero, therefore (x+1) is not a factor x4 – x3 + x2 – x + 1



Question 3.

Determine which of the following polynomials has (x + 1) as a factor.

x4 + 2x3 + 2x2 + x + 1


Answer:

Let f(x) = x4 + 2x3 + 2x2 + x + 1


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = (–1)4 + 2(–1)3 + 2(–1)2 + (–1) + 1


⇒ f(–1) = 1 – 2 + 2 – 1 + 1


⇒ f(–1) = 1


As, f(–1) is not equal to zero, therefore (x+1) is not a factor x4 + 2x3 + 2x2 + x + 1



Question 4.

Determine which of the following polynomials has (x + 1) as a factor.

x3 – x2 –(3 – √3 ) x + √3


Answer:

Let f(x) = x3 – x2 – (3 –√3 ) x + √3


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = (–1)3 – (–1)2 – (3 –√3)(–1) + √3


⇒ f(–1) = –1 – 1 + 3 – √3 + √3


⇒ f(–1) = 1


As, f(–1) is not equal to zero, therefore (x+1) is not a factor x3 – x2 –(3 –√3) x + √3



Question 5.

Use the Factor Theorem to determine whether g(x) is factor of f(x) in the following cases:

f(x) = 5x3 + x2 – 5x – 1, g(x) = x + 1


Answer:

By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = 5 (–1)3 + (–1)2 – 5(–1) – 1


⇒ f(–1) = –5 + 1 + 5 – 1


⇒ f(–1) = 0


As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)



Question 6.

Use the Factor Theorem to determine whether g(x) is factor of f(x) in the following cases:

f(x) = x3 + 3x2 + 3x + 1, g(x) = x + 1


Answer:

By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


⇒ f(–1) = (–1)3 + 3(–1)2 + 3(–1) + 1


⇒ f(–1) = – 1 + 3 – 3 + 1


⇒ f(–1) = 0


As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)



Question 7.

Use the Factor Theorem to determine whether g(x) is factor of f(x) in the following cases:

f(x) = x3 – 4x2 + x + 6, g(x) = x – 2


Answer:

By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0


For checking (x – 2) to be a factor, we will find f(2)


⇒ f(2) = (2)3 – 4(2)2 + (2) + 6


⇒ f(–1) = 8 – 16 + 2 + 6


⇒ f(–1) = 0


As, f(–1) is equal to zero, therefore, g(x) = (x – 2) is a factor of f(x)



Question 8.

Use the Factor Theorem to determine whether g(x) is factor of f(x) in the following cases:

f(x) = 3x3 + x2 – 20x + 12, g(x) = 3x–2


Answer:

By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0


For checking (3x – 2) to be a factor, we will find f(2/3)





As, f(–1) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)



Question 9.

Use the Factor Theorem to determine whether g(x) is factor of f(x) in the following cases:

f(x) = 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3


Answer:

By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0


For checking (2x + 3) to be a factor, we will find f(–3/2)





As, f(–3/2) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)



Question 10.

Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24.


Answer:

Let f(x) = x3 – 3x2 – 10x + 24


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0


For checking (x – 2) to be a factor, we will find f(2)


⇒ f(2) = (2)3 – 3(2)2 – 10(2) + 24


⇒ f(2) = 8 – 12 – 20 + 24


⇒ f(2) = 0


So, (x–2) is a factor.


For checking (x + 3) to be a factor, we will find f(–3)


⇒ f(–3) = (–3)3 – 3(–3)2 – 10(–3) + 24


⇒ f(–3) = –27 – 27 + 30 + 24


⇒ f(–3) = 0


So, (x+3) is a factor.


For checking (x – 4) to be a factor, we will find f(4)


⇒ f(4) = (4)3 – 3(4)2 – 10(4) + 24


⇒ f(4) = 64 – 48 – 40 + 24


⇒ f(4) = 0


So, (x–4) is a factor.


∴ (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24



Question 11.

Show that (x + 4), (x – 3) and (x – 7) are factors of x3 – 6x2 – 19x + 84.


Answer:

Let f(x) = x3 – 6x2 – 19x + 84


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0


For checking (x + 4) to be a factor, we will find f(–4)


⇒ f(–4) = (–4)3 – 6(–4)2 – 19(–4) + 84


⇒ f(–4) = –64 – 96 + 76 + 84


⇒ f(–4) = 0


So, (x+4) is a factor.


For checking (x – 3) to be a factor, we will find f(3)


⇒ f(3) = (3)3 – 6(3)2 – 19(3) + 84


⇒ f(3) = 27 – 54 – 57 + 84


⇒ f(3) = 0


So, (x–3) is a factor.


For checking (x – 7) to be a factor, we will find f(7)


⇒ f(7) = (7)3 – 6(7)2 – 19(7) + 84


⇒ f(7) = 343 – 294 – 133 + 84


⇒ f(7) = 0


So, (x–7) is a factor.


∴ (x + 4), (x – 3) and (x – 7) are factors of x3 – 3x2 – 10x + 24



Question 12.

If both (x – 2) and  are factors of px2 + 5x + r, show that p = r.


Answer:

Let f(x) = px2 + 5x + r


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.


So, if (x – 2) is a factor of f(x)


⇒ f(2) = 0


⇒ p(2)2 + 5(2) + r = 0


⇒ 4p + r = –10 -------- (A)


Also as  is also a factor,





⇒ p + 4r = –10 ----- (B)

Subtract B from A to get,


4p + r - (p + 4r)= –10 - (-10)

4p + r - p - 4r = -10 + 10

3p - 3r = 0

3p = 3r

p = r

Question 13.

If (x2 – 1) is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0


Answer:

Let f(x) = ax4 + bx3 + cx2 + dx + e


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.


Also we can write, (x2 – 1) = (x + 1)(x – 1)


Since (x2 – 1) is a factor of f(x), this means (x + 1) and (x – 1) both are factors of f(x).


So, if (x – 1) is a factor of f(x)


⇒ f(1) = 0


⇒ a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0


⇒ a + b + c + d + e = 0 ----- (A)


Also as (x + 1) is also a factor,


⇒ f(–1) = 0


⇒ a(–1)4 + b(–1)3 + c(–1)2 + d(–1) + e = 0


⇒ a – b + c – d + e = 0


⇒ a + c + e = b + d ---- (B)


On solving equations (A) and (B), we get,


a + c + e = b + d = 0



Question 14.

Factorize

x3 – 2x2 – x + 2


Answer:

Let p(x) = x3 – 2x2 – x + 2


By trial, we find that p(1) = 0, so by Factor theorem,


(x – 1) is the factor of p(x)


When we divide p(x) by (x – 1), we get x2 – x – 2.


Now, (x2 – x – 2) is a quadratic and can be solved by splitting the middle terms.


We have x2 – x – 2 = x2 – 2x + x – 2


⇒ x (x – 2) + 1 (x – 2)


⇒ (x + 1)(x – 2)


So, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)



Question 15.

Factorize

x3 – 3x2 – 9x – 5


Answer:

Let p(x) = x3 – 3x2 – 9x – 5


By trial, we find that p(–1) = 0, so by Factor theorem,


(x + 1) is the factor of p(x)


When we divide p(x) by (x + 1), we get x2 – 4x – 5.


Now, (x2 – 4x – 5) is a quadratic and can be solved by splitting the middle terms.


We have x2 – 4x – 5 = x2 – 5x + x – 5


⇒ x (x – 5) + 1 (x – 5)


⇒ (x + 1)(x – 5)


So, x3 – 3x2 – 9x – 5= (x + 1)(x + 1)(x – 5)



Question 16.

Factorize

x3 + 13x2 + 32x + 20


Answer:

Let p(x) = x3 + 13x2 + 32x + 20


By trial, we find that p(–1) = 0, so by Factor theorem,


(x + 1) is the factor of p(x)


When we divide p(x) by (x + 1), we get x2 + 12x + 20.


Now, (x2 + 12x + 20) is a quadratic and can be solved by splitting the middle terms.


We have x2 + 12x + 20= x2 + 10x + 2x + 20


⇒ x (x + 10) + 2 (x + 10)


⇒ (x + 2)(x + 10)


So, x3 + 13x2 + 32x + 20 = (x + 1)(x + 2)(x + 10)


Question 17.

Factorize

y3 + y2 – y – 1


Answer:

Let p(y) = y3 + y2 – y – 1


On taking y2 common from first two terms in p(y), we get,


p (y) = y2(y + 1) –1(y + 1)


Now, taking (y + 1) common, we get,


⇒ p(y) = (y2 – 1)(y + 1)


As we know the identity, (y2 – 1) = (y + 1)(y – 1)


⇒ p(y) = (y – 1)(y + 1)(y + 1)



Question 18.

If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that c = 0 and a = b.


Answer:

Let f(x) = ax2 + bx + c and p(x) = bx2 + ax + c


As (x + 1) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.


⇒ f(–1) = p (–1) = 0


⇒ a(–1)2 + b(–1) + c = b(–1)2 + a(–1) + c


⇒ a – b + c = b – a + c


⇒ 2a = 2b


⇒ a = b ------ (A)


Also, we discussed that,


f(–1) = 0


⇒ a(–1)2 + b(–1) + c = 0


⇒ a – b + c = 0


From equation (A), we see that a = b,


⇒ c = 0 -------- (B)


∴ Equations (A) and (B) show us the required result.



Question 19.

If x2 – x – 6 and x2 + 3x – 18 have a common factor (x – a) then find the value of a.


Answer:

Let f(x) = x2 – x – 6 and p(x) = x2 + 3x – 18


As (x – a) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.


⇒ f(a) = p (a)


⇒ (a)2 – (a) – 6 = (a)2 + 3(a) – 18


⇒ 4a = 12


⇒ a = 3


∴ The value of a is 3.



Question 20.

If (y – 3) is a factor of y3 – 2y2 – 9y + 18 then find the other two factors.


Answer:

Let f(x) = y3 – 2y2 – 9y + 18


Taking y2 common from the first two terms of f(x) and 9 from the last two terms of f(x), we get,


⇒ f(x) = y2(y – 2) –9(y – 2)


Now, taking (y – 2) common from above,


⇒ f(x) = (y2 – 9)(y – 2) -------- (A)


We know the identity as,


a2 – b2 = (a – b)(a + b)


So, using above identity on equation (A), we get,


⇒ f(x) = (y + 3)(y – 3)(y – 2)


∴ the other two factors of y3 – 2y2 – 9y + 18 besides (y – 3) are (y + 3) and (y – 2).




Exercise 2.5
Question 1.

Use suitable identities to find the following products

(x + 5) (x + 2)


Answer:

using the identity (x + a) × (x + b) = x2 + (a + b)x + ab


here a = 5 and b = 2


⇒ (x + 5) (x + 2) = x2 + (5 + 2)x + 5 × 2


Therefore (x + 5) (x + 2) = x2 + 7x + 10



Question 2.

Use suitable identities to find the following products

(x - 5) (x - 5)


Answer:

(x - 5) (x - 5) = (x – 5)2


Using identity (a – b)2 = a2 – 2ab + b2


Here a = x and b = 5


⇒ (x - 5) (x - 5) = x2 – 2 × x × 5 + 52


⇒ (x - 5) (x - 5) = x2 – 10x + 25


Therefore (x - 5) (x - 5) = x2 – 10x + 25



Question 3.

Use suitable identities to find the following products

(3x + 2)(3x - 2)


Answer:

using the identity (a + b) × (a – b) = a2 – b2


here a = 3x and b = 2


⇒ (3x + 2)(3x - 2) = (3x)2 – 22


Therefore (3x + 2)(3x - 2) = 9x2 – 4



Question 4.

Use suitable identities to find the following products



Answer:

using the identity (a + b) × (a – b) = a2 – b2


here a = x2 and b = 


⇒  = (x2)2 - 


Therefore  = x4 - 



Question 5.

Use suitable identities to find the following products

(1 + x) (1 + x)


Answer:

(1 + x) (1 + x) = (1 + x)2


Using identity (a + b)2 = a2 + 2ab + b2


Here a = 1 and b = x


⇒ (1 + x) (1 + x) = 12 + 2(1)(x) + x2


Therefore (1 + x) (1 + x) = 1 + 2x + x2



Question 6.

Evaluate the following products without actual multiplication.

101 × 99


Answer:

101 can be written as (100 + 1) and


99 can be written as (100 - 1)


⇒ 101 × 99 = (100 + 1) × (100 - 1)


using the identity (a + b) × (a – b) = a2 – b2


here a = 100 and b = 1


⇒ 101 × 99 = 1002 – 12


⇒ 101 × 99 = 10000 – 1


⇒ 101 × 99 = 9999



Question 7.

Evaluate the following products without actual multiplication.

999 × 999


Answer:

999 can be written as (1000 – 1)


⇒ 999 × 999 = (1000 – 1) × (1000 – 1)


⇒ 999 × 999 = (1000 – 1)2


Using identity (a – b)2 = a2 – 2ab + b2


Here a = 1000 and b = 1


⇒ 999 × 999 = 10002 – 2(1000)(1) + 12


⇒ 999 × 999 = 1000000 – 2000 + 1


⇒ 999 × 999 = 998000 + 1


⇒ 999 × 999 = 998001



Question 8.

Evaluate the following products without actual multiplication.



Answer:

⇒ 50 =  and 49 = 


⇒ 50 =  and 49 = 


⇒ 50 =  and 49 = 


⇒  =  × 


⇒  = 


Consider 101 × 99


101 can be written as (100 + 1) and


99 can be written as (100 - 1)


⇒ 101 × 99 = (100 + 1) × (100 - 1)


using the identity (a + b) × (a – b) = a2 – b2


here a = 100 and b = 1


⇒ 101 × 99 = 1002 – 12


⇒ 101 × 99 = 10000 – 1


⇒ 101 × 99 = 9999


Therefore  = 



Question 9.

Evaluate the following products without actual multiplication.

501 × 501


Answer:

501 can be written as (500 + 1)


⇒ 501 × 501 = (500 + 1) × (500 + 1)


⇒ 501 × 501 = (500 + 1)2


⇒ 501 × 501 = (500 + 1) × (500 + 1)


⇒ 501 × 501 = (500 + 1)2


Using identity (a + b)2 = a2 + 2ab + b2


Here a = 500 and b = 1


⇒ 501 × 501 = 5002 + 2(500)(1) + 12


⇒ 501 × 501 = 250000 + 1000 + 1


⇒ 501 × 501 = 251001



Question 10.

Evaluate the following products without actual multiplication.

30.5 × 29.5


Answer:

30.5 =  and 29.5 = 


⇒ 30.5 × 29.5 =  × 


⇒ 30.5 × 29.5 =  × 


⇒ 30.5 × 29.5 =  …(i)


Consider 61 × 59


61 = (60 + 1)


59 = (60 – 1)


⇒ 61 × 59 = (60 + 1)(60 – 1)


using the identity (a + b) × (a – b) = a2 – b2


here a = 60 and b = 1


⇒ 61 × 59 = 602 – 12


⇒ 61 × 59 = 3600 – 1


⇒ 61 × 59 = 3599


From (i)


⇒ 30.5 × 29.5 = 


Therefore 30.5 × 29.5 = 899.75



Question 11.

Factorise the following using appropriate identities.

16x2 + 24xy + 9y2


Answer:

16x2 can be written as (4x)2


24xy can be written as 2(4x)(3y)


9y2 can be written as (3y)2


⇒ 16x2 + 24xy + 9y2 = (4x)2 + 2(4x)(3y) + (3y)2


Using identity (a + b)2 = a2 + 2ab + b2


Here a = 4x and b = 3y


⇒ 16x2 + 24xy + 9y2 = (4x + 3y)2


Therefore 16x2 + 24xy + 9y2 = (4x + 3y) (4x + 3y)



Question 12.

Factorise the following using appropriate identities.

4y2 - 4y + 1


Answer:

4y2 can be written as (2y)2


4y can be written as 2(1)(2y)


1 can be written as 12


⇒ 4y2 - 4y + 1 = (2y)2 - 2(1)(2y) + 12


Using identity (a - b)2 = a2 - 2ab + b2


Here a = 2y and b = 1


⇒ 4y2 - 4y + 1 = (2y - 1)2


Therefore 4y2 - 4y + 1 = (2y - 1) (2y - 1)



Question 13.

Factorise the following using appropriate identities.



Answer:

4x2 can be written as (2x)2


 can be written as 


⇒ 4x2 -  = (2x)2 - 


using the identity (a + b) × (a – b) = a2 – b2


here a = 2x and b = 


Therefore 4x2 -  = (2x + ) (2x - )


Question 14.

Factorise the following using appropriate identities.

18a2 – 50


Answer:

Take out common factor 2


⇒ 18a2 – 50 = 2 (9a2 - 25)


Now


9a2 can be written as (3a)2


25 can be written as 52


⇒ 18a2 – 50 = 2 ((3a)2 – 52)


using the identity (a + b) × (a – b) = a2 – b2


here a = 3a and b = 5


therefore 18a2 – 50 = 2 (3a + 5) (3a – 5)



Question 15.

Factorise the following using appropriate identities.

x2 + 5x + 6


Answer:

Given is quadratic equation which can be factorised by splitting the middle term as shown


⇒ x2 + 5x + 6 = x2 + 3x + 2x + 6


= x (x + 3) + 2 (x + 3)


= (x + 3) (x + 2)


Therefore x2 + 5x + 6 = (x + 3) (x + 2)



Question 16.

Factorise the following using appropriate identities.

3p2 - 24p + 36


Answer:

Take out common factor 3


⇒ 3p2 - 24p + 36 = 3 (p2 – 8p + 12)


Now splitting the middle term of quadratic p2 – 8p + 12 to factorise it


⇒ 3p2 - 24p + 36 = 3 (p2 – 6p – 2p + 12)


= 3 [p (p – 6) – 2 (p – 6)]


= 3 (p – 2) (p – 6)



Question 17.

Expand each of the following, using suitable identities

(x + 2y + 4z)2


Answer:

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca


Here a = x, b = 2y and c = 4z


⇒ (x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)


Therefore


(x + 2y + 4z)2 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz



Question 18.

Expand each of the following, using suitable identities

(2a - 3b)3


Answer:

Using identity (x – y)3 = x3 - y3 – 3x2y + 3xy2


Here x = 2a and y = 3b


⇒ (2a - 3b)3 = (2a)3 – (3b)3 – 3(2a)2(3b) + 3(2a)(3b)2


= 8a3 – 27b3 – 18a2b + 18ab2


Therefore (2a - 3b)3 = 8a3 – 27b3 – 18a2b + 18ab2



Question 19.

Expand each of the following, using suitable identities

(-2a + 5b - 3c)2


Answer:

(-2a + 5b - 3c)2 = [(-2a) + (5b) + (-3c)]2


Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


Here x = -2a, y = 5b and z = -3c


⇒ (-2a + 5b - 3c)2 = (-2a)2 + (5b)2 + (-3c)2 + 2(-2a)(5b) + 2(5b)(-3c) + 2(-3c)(-2a)


⇒ (-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 + (-20ab) + (-30bc) + 12ac


⇒ (-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12ac


Therefore


(-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12ac



Question 20.

Expand each of the following, using suitable identities



Answer:

 = [ +  + 1]2


Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


Here x = , y =  and z = 1


⇒  =  +  + 12 + 2 + 2(1) + 2(1)


⇒  =  +  + 1 +  + (-b) + 


⇒  =  +  + 1 -  – b + 


Therefore  =  +  + 1 -  – b + 



Question 21.

Expand each of the following, using suitable identities

(p + 1)3


Answer:

Using identity (x + y)3 = x3 + y3 + 3x2y + 3xy2


Here x = p and y = 1


⇒ (p + 1)3 = p3 + 13 + 3(p)2(1) + 3(p)(1)2


= p3 + 13 + 3p2 + 3p


Therefore (p + 1)3 = p3 + 13 + 3p2 + 3p



Question 22.

Expand each of the following, using suitable identities



Answer:

Using identity (a – b)3 = a3 - b3 – 3a2b + 3ab2


Here a = x and b = y


⇒  = x3 -  – 3(x)2() + 3(x)


⇒  = x3 -  – 2x2y + xy2


Therefore  = x3 -  – 2x2y + xy2



Question 23.

Factorise

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz


Answer:

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz


25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = 25x2 + 16y2 + 4z2 + (-40xy) + 16yz + (-20xz)


25x2 can be written as (-5x)2


16y2 can be written as (4y)2


4z2 can be written as (2z)2


-40xy can be written as 2(-5x)(4y)


16yz can be written as 2(4y)(2z)


-20xz can be written as 2(-5x)(2z)


⇒ 25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (-5x)2 + (4y)2 +


(2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) …(i)


Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca


Comparing (-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) with a2 + b2 + c2 + 2ab + 2bc + 2ca we get


a = -5x, b = 4y and c = 2z


therefore


(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) = (- 5x + 4y + 2z)2


From (i)


25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (- 5x + 4y + 2z)2



Question 24.

Factorise

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca


Answer:

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca


9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = 9a2 + 4b2 + 16c2 + 12ab + (-16bc) + (-24ca)


9a2 can be written as (3a)2


4b2 can be written as (2b)2


16c2 can be written as (-4c)2


12ab can be written as 2(3a)(2b)


-16bc can be written as 2(2b)(-4c)


-24ca can be written as 2(-4c)(3a)


⇒ 9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)


Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


Comparing (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x2 + y2 + z2 + 2xy + 2yz + 2zx we get


x = 3a, y = 2b and z = -4c


therefore


(3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) = (3a + 2b + (-4c))2


From (i)


9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a + 2b – 4c)2


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