Polynomials Class 10th Mathematics AP Board Solution

Class 10th Mathematics AP Board Solution
Exercise 3.1
  1. If p(x) = 5x^7 - 6x^5 + 7x - 6, find (i) coefficient of x^5 (ii) degree of p(x)…
  2. State which of the following statements are true and which are false? Give…
  3. If p(t) t^3 - 1, find the values of p(1),p(-1),P(0),p(2),p(-2).
  4. Check whether -2 and 2 are the zeroes of the polynomial x^4 - 16
  5. Check whether 3 and -2 are the zeroes of the polynomial when p(x) = x^2 - x - 6…
Exercise 3.2
  1. The graphs of y = p(x) are given in the figure below, for some polynomials p(x)…
  2. p(x) = 3x Find the zeroes of the given polynomials.
  3. p(x) = x^2 + 5x + 6 Find the zeroes of the given polynomials.
  4. p(x)- (x+2)(x+3) Find the zeroes of the given polynomials.
  5. p(x)=x^4 - 16 Find the zeroes of the given polynomials.
  6. p(x) = x^2 - x - 12 Draw the graphs of the given polynomial and find the…
  7. p(x) = x^2 - 6x + 9 Draw the graphs of the given polynomial and find the…
  8. p(x) = x^2 - 4x + 5 Draw the graphs of the given polynomial and find the…
  9. p(x) = x^2 + 3x - 4 Draw the graphs of the given polynomial and find the…
  10. p(x) = x^2 - 1 Draw the graphs of the given polynomial and find the zeroes.…
  11. Why are 1/4 and -1 zeroes of the polynomials p(x) - 4x^2 + 3x - 1?…
Exercise 3.3
  1. x^2 - 2x - 8 Find the zeroes of the following quadratic polynomials and verify…
  2. 4s^2 - 4s + 1 Find the zeroes of the following quadratic polynomials and verify…
  3. 6x^2 - 3 - 7x Find the zeroes of the following quadratic polynomials and verify…
  4. 4u^2 + 8u Find the zeroes of the following quadratic polynomials and verify the…
  5. t^2 - 15 Find the zeroes of the following quadratic polynomials and verify the…
  6. 3x^2 - x - 4 Find the zeroes of the following quadratic polynomials and verify…
  7. 1/4, -1 Find the quadratic polynomial in each case, with the given numbers as…
  8. root 2 , 1/3 Find the quadratic polynomial in each case, with the given numbers…
  9. 0 , root 5 Find the quadratic polynomial in each case, with the given numbers…
  10. 1, 1 Find the quadratic polynomial in each case, with the given numbers as the…
  11. 1/4, 1/4 Find the quadratic polynomial in each case, with the given numbers as…
  12. 4, 1 Find the quadratic polynomial in each case, with the given numbers as the…
  13. 2,-1 Find the quadratic polynomial, for the zeroes given in each case.…
  14. root 3 , root 3 Find the quadratic polynomial, for the zeroes given in each…
  15. 1/4, -1 Find the quadratic polynomial, for the zeroes given in each case.…
  16. 1/2, 3/2 Find the quadratic polynomial, for the zeroes given in each case.…
  17. Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x^3 + 3x^2 - x -…
Exercise 3.4
  1. p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2 Divide the polynomial p(x) by the…
  2. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and…
  3. p(x) = x^4 - 5x + 6, g(x) = 2 - x^2 Divide the polynomial p(x) by the…
  4. t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12 Check in which case the first polynomial…
  5. x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2 Check in which case the first…
  6. x^2 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1 Check in which case the first…
  7. Obtain all other zeroes of 3x^4 + 6x^3 - 2x^2 - 10x - 5, if two of its zeroes…
  8. On dividing x^3 - 3x^2 + x + 2 by a polynomial g(x), the quotient and remainder…
  9. deg p (x) = deg q (x) Give examples of polynomials p(x), g(x), q(x) and r(x)…
  10. deg q (x) = deg r (x) Give examples of polynomials p(x), g(x), q(x) and r(x)…
  11. deg r (x) = 0 Give examples of polynomials p(x), g(x), q(x) and r(x) which…

Exercise 3.1
Question 1.

If p(x) = 5x7 – 6x5 + 7x - 6, find

(i) coefficient of x5

(ii) degree of p(x)

(iii) constant term.


Answer:

(i) A coefficient is a multiplicative factor in some term of a polynomial. It is the constant (-6 here) written before the variable (x5 here).

∴ The coefficient of x5 is -6.


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


∴ The degree of p(x) is 7.


(iii) The term that is not attached to a variable (i.e., x) is -6.


∴ The constant term is -6.



Question 2.

State which of the following statements are true and which are false? Give reasons for your choice.

(i) The degree of the polynomial  is 

(ii) The coefficient of x2 in the polynomial p(x) = 3x3 – 4x2 + 5x + 7 is 2.

(iii) The degree of a constant term is zero.

(iv)  is a quadratic polynomial.

(v) The degree of a polynomial is one more than the number of terms in it.


Answer:

(i) False

Since, Degree of a polynomial is the highest power of x in the polynomial which is 2 in.


(ii) False


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


∴ The coefficient of x2 is -4.


(iii) True


Since, the power of x of a constant term is 0.


∴ the degree of a constant term is 0.


(iv) False


For an expression to be a polynomial term, any variables in the expression must have whole-number powers (i.e., x0, x1, x2,……)


Since, the power in the expression  has powers -2 and -1 which are not whole numbers, therefore, is not a polynomial.


(v) False


Degree of a polynomial is the highest power of x in the polynomial. The degree of polynomial is not related to the number of terms in the polynomial.


Therefore, the statement is false.



Question 3.

If p(t) t3 – 1, find the values of p(1),p(-1),P(0),p(2),p(-2).


Answer:

P(t) = t3 – 1

Therefore,


P(1) = 13 – 1


⇒ P(1) = 1 – 1


⇒ P(1) = 0


P(-1) = (-1)3 – 1


⇒ P(-1) = -1 – 1


⇒ P(-1) = -2


P(0) = 03 – 1


⇒ P(0) = 0 – 1


⇒ P(0) = -1


P(2) = 23 – 1


⇒ P(2) = 8 – 1


⇒ P(2) = 7


P(-2) = (-2)3 – 1


⇒ P(2) = -8 – 1


⇒ P(2) = -9



Question 4.

Check whether -2 and 2 are the zeroes of the polynomial x4 - 16


Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.


Now,


P(x) = x4 – 16


Therefore,


P(-2) = (-2)4 – 16


⇒ P(-2) = 16 – 16


⇒ P(-2) = 0


P(2) = 24 – 16


⇒ P(2) = 16 – 16


⇒ P(2) = 0


Hence, Yes, -2 and -2 are zeroes of the polynomial x4 – 16.



Question 5.

Check whether 3 and -2 are the zeroes of the polynomial when p(x) = x2 – x - 6


Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.


Now,


P(x) = x2 – x – 6


Therefore,


P(-2) = (-2)2 – (-2) – 6


⇒ P(-2) = 4 + 2 – 6


⇒ P(-2) = 0


P(3) = 32 – 3 – 6


⇒ P(3) = 9 – 3 – 6


-⇒ P(3) = 0


Hence, Yes, 3 and -2 are zeroes of the polynomial x2 – x – 6.




Exercise 3.2
Question 1.

The graphs of y = p(x) are given in the figure below, for some polynomials p(x) In each case, find the number of zeroes of p(x)



Answer:

(i) Since, the graph does not intersect with x-axis at any point therefore, it has no zeroes.

(ii) Since, the graph intersects with x-axis at only one point therefore, it has 1 number of zeroes.


(iii) Since, the graph intersects with x-axis three points therefore, it has 3 number of zeroes.


(iv) Since, the graph intersects with x-axis two points therefore, it has 2 number of zeroes.


(v) Since, the graph intersects with x-axis at four points therefore, it has 4 number of zeroes.


(vi) Since, the graph intersects with x-axis at three points therefore, it has 3 number of zeroes.



Question 2.

Find the zeroes of the given polynomials.

p(x) = 3x


Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, to find zeroes put p(x)=0.


p(x) = 0


⇒ 3x = 0


⇒ x = 0


Hence, x=0 is the zero of the polynomial.



Question 3.

Find the zeroes of the given polynomials.

p(x) = x2 + 5x + 6


Answer:

p(x) = 0

⇒ x2 + 5x + 6 = 0


⇒ x2 + 3x + 2x + 6 = 0


⇒ x(x + 3) + 2(x + 3) = 0


⇒ (x + 3)(x + 2) = 0


⇒ x = -3 and x = -2


Hence, x=-3 and x=-2 are the zeroes of the polynomial.



Question 4.

Find the zeroes of the given polynomials.

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


Answer:

p(x) = 0

⇒ (x + 3)(x + 2)


⇒ x = -3 and x = -2


Hence, x=-3 and x=-2 are the zeroes of the polynomial.



Question 5.

Find the zeroes of the given polynomials.

p(x)=x4 - 16


Answer:

p(x) = 0

⇒ x4 – 16 = 0


⇒ (x2 – 4)(x2 + 4)= 0


⇒ (x + 2)(x – 2)(x2 + 4)= 0


⇒ x = -2, x = 2 and x2 = 4


⇒ x = -2, x = 2 and x = ±2


Hence, x=2 and x=-2 are the zeroes of the polynomial.



Question 6.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x2 – x - 12


Answer:

p(x) = x2 – x – 12



Clearly, the graph intersects with x – axis at x = -3 and x = 4


Hence, the zeroes of the polynomial x2 – x – 12 are x = -3 and x=4.



Question 7.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x2 – 6x + 9


Answer:

p(x) = x2 – 6x + 9



Clearly, the graph intersects with x – axis at x = 4


Hence, the zeroes of the polynomial x2 – 6x + 9 are x=3.



Question 8.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x2 – 4x + 5


Answer:

p(x) = x2 – 4x + 5



Clearly, the graph does not intersect with x – axis at any point.


Hence, there are no zeroes of the polynomial x2 – 4x + 5.



Question 9.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x2 + 3x - 4


Answer:

p(x) = x2 + 3x – 4



Clearly, the graph intersects with x – axis at x = -4 and x = 1.


Hence, the zeroes of the polynomial x2 + 3x – 4 are x = -4 and x= 1



Question 10.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x2 - 1


Answer:

p(x) = x2 – 1



Clearly, the graph intersects with x – axis at x = -1 and x = 1.


Hence, the zeroes of the polynomial x2 – 1 are x = -1 and x= 1



Question 11.

Why are 1/4 and -1 zeroes of the polynomials p(x) – 4x2 + 3x – 1?


Answer:


A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, if p(x)=0, for a given x then the value of x is zero of polynomial.


p(x) = 4x2 + 3x – 1


For 1/4,


p(1/4) = 4(1/4)2 + 3(1/4) – 1


⇒ p(1/4) = 1/4 + 3/4 – 1


⇒ p(1/4) = 1 – 1


⇒ p(1/4) = 0


Therefore, x = 1/4 is a zero of the polynomial 4x2 + 3x – 1.


For -1,


p(-1) = 4(-1)2 + 3(-1) – 1


⇒ p(-1) = 4 – 3 – 1


⇒ p(-1) = 4 – 4


⇒ p(-1) = 0


Therefore, x = -1 is a zero of the polynomial 4x2 + 3x – 1.


Hence, x = 1/4 and x = -1 are zeroes of the polynomial 4x2 + 3x – 1.




Exercise 3.3
Question 1.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

x2 – 2x – 8


Answer:

p(x) = 0

⇒ x2 – 2x – 8 = 0


⇒ x2 + 2x – 4x – 8 = 0


⇒ x(x + 2) – 4(x + 2) = 0


⇒ (x + 2)(x – 4) = 0


⇒ x = -2 and x = 4


Hence, -2 and 4 are zeroes of the polynomial x2 – 2x – 8.


Now,


Sum of zeroes = -2 + 4


⇒ Sum of zeroes = 2





Product of zeroes = -2 × 4


⇒ Product of zeroes = -8






Question 2.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

4s2 – 4s + 1


Answer:

p(s) = 0

⇒ 4s2 – 4s + 1 = 0


⇒ 4s2 – 2s – 2s + 1 = 0


⇒ 2s(2s – 1) – 1(2s – 1) = 0


⇒ (2s – 1) (2s – 1) = 0


⇒ x = 1/2 and x = 1/2


Hence, 1/2 and 1/2 are zeroes of the polynomial 4s2 – 4s + 1.


Now,


Sum of zeroes = 1/2 + 1/2


⇒ Sum of zeroes = 1





Product of zeroes = 1/2 × 1/2


⇒ Product of zeroes = 1/4





Question 3.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

6x2 – 3 – 7x


Answer:

p(x) = 0

⇒ 6x2 – 3 – 7x = 0


⇒ 6x2 – 7x – 3 = 0


⇒ 6x2 – 9x + 2x – 3 = 0


⇒ 3x(2x – 3) + 1(2x – 3) = 0


⇒ (2x – 3) (3x + 1) = 0


⇒  and 


Hence,  and  are zeroes of the polynomial 6x2 – 3 – 7x.


Now,









⇒ Product of zeroes = -1/2






Question 4.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

4u2 + 8u


Answer:

p(u) = 0

⇒ 4u2 + 8u = 0


⇒ 4u(u + 2) = 0


⇒ u = 0 and u = -2


Hence, -2 and 0 are zeroes of the polynomial 4u2 + 8u.


Now,


Sum of zeroes = -2 + 0


⇒ Sum of zeroes = -2





Product of zeroes = -2 × 0


⇒ Product of zeroes = 0






Question 5.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

t2 - 15


Answer:

p(t) = 0

⇒ t2 – 15 = 0


⇒ (t – √15) (t + √15) = 0


⇒ t = -√15 and t = √15


Hence, -√15 and √15 are zeroes of the polynomial t2 – 15.


Now,


Sum of zeroes = -√15 + √15


⇒ Sum of zeroes = -√15 + √15


⇒ Sum of zeroes = 0





Product of zeroes = -√15 × √15


⇒ Product of zeroes = -15






Question 6.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

3x2 – x – 4


Answer:

p(x) = 0

⇒ 3x2 – x – 4 = 0


⇒ 3x2 + 3x – 4x – 4 = 0


⇒ 3x(x + 1) –4(x + 1) = 0


⇒ (x + 1)(3x – 4)= 0



Hence, -1 and are zeroes of the polynomial 3x2 – x – 4.


Now,














Question 7.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1/4, -1


Answer:

Given: α + β = 1/4

αβ = -1


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β .


Now we know that,



 …(2)


And,



If a = 4


⇒ c = -4 …(3)


From (2) and (3),


a = 4, b = -1 and c = -4


Hence, the polynomial is 4x2 – x – 4



Question 8.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.



Answer:

Given: α + β = √2


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β .


Now we know that,




And,



If a = 3



⇒ b = -3√2 …(2)


⇒c = 1 …(3)


From (2) and (3),


a = 3, b = -3 √2 and c = 1


Hence, the polynomial is 3x2 – 3√2x + 1



Question 9.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.



Answer:

Given: α + β = 0

αβ = √5


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β .


Now we know that,



 …(2)


And,



If a = 1


⇒c = √5 …(3)


From (2) and (3),


a = 1, b = 0 and c = √5


Hence, the polynomial is x2 +√5



Question 10.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1, 1


Answer:

Given: α + β = 1

αβ = 1


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β .


Now we know that,




And,



If a = 1



⇒ b = -1 …(2)


⇒c = 1 …(3)


From (2) and (3),


a = 1, b = -1 and c = 1


Hence, the polynomial is x2 – x + 1



Question 11.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1/4, 1/4


Answer:

Given: α + β = -1/4

αβ = 1/4


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β .


Now we know that,



And,



If a = 4



⇒ b = 1 …(2)


⇒c = 1 …(3)


From (2) and (3),


a = 4, b = 1 and c = 1


Hence, the polynomial is 4x2 + x + 1



Question 12.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

4, 1


Answer:

Given: α + β = 4

αβ = 1


Let the quadratic polynomial be ax2 + bx + c …(1)


Where, a≠0


And zeroes of the polynomial are α and β.


Now we know that,



And,





If a = 1



⇒ b = -4 …(2)


⇒c = 1 …(3)


From (2) and (3),


a = 1, b = -4 and c = 1


Hence, the polynomial is x2 - 4x + 1



Question 13.

Find the quadratic polynomial, for the zeroes given in each case.

2,-1


Answer:

Let the quadratic polynomial be ax2 + bx + c

And, its zeroes be α and β


α = 2


β = -1








If a = 1,



⇒ b = -1



⇒ c = -2


Hence, the polynomial is x2 – x – 2



Question 14.

Find the quadratic polynomial, for the zeroes given in each case.



Answer:

Let the quadratic polynomial be ax2 + bx + c

And, its zeroes be α and β


α = √3


β = -√3








If a = 1,



⇒ b = 0



⇒ c = -3


Hence, the polynomial is x2 – 3



Question 15.

Find the quadratic polynomial, for the zeroes given in each case.

1/4, -1


Answer:

Let the quadratic polynomial be ax2 + bx + c

And, its zeroes be α and β


α = 1/4


β = -1








If a = 4,



⇒ b = 3



⇒ c = -1


Hence, the polynomial is 4x2 + 3x – 1.



Question 16.

Find the quadratic polynomial, for the zeroes given in each case.

1/2, 3/2


Answer:

Let the quadratic polynomial be ax2 + bx + c

And, its zeroes be α and β










If a = 4,



⇒ b = -8



⇒ c = 3


Hence, the polynomial is 4x2 – 8x + 3.



Question 17.

Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x3 + 3x2 – x - 3 and check the relationship between zeroes and the coefficients.


Answer:

P(x) = x3 + 3x2 – x – 3

For x = 1,


⇒ P(1) = 13 + 3(1)2 – 1 – 3


⇒ P(1) = 1+ 3 – 1 – 3


⇒ P(1) = 0


For x = -1,


⇒ P(-1) = (-1)3 + 3(-1)2 – (-1) – 3


⇒ P(-1) = -1+ 3 + 1 – 3


⇒ P(-1) = 0


For x = -3,


⇒ P(-3) = (-3)3 + 3(-3)2 – (-3) – 3


⇒ P(-3) = -27+ 27 + 3 – 3


⇒ P(-3) = 0


Now,



Sum of zeroes = 1 + (-1) + (-3) = -3


Hence,



And,



Product of zeroes = 1 × (-1) × (-3) = -3





Exercise 3.4
Question 1.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 - 2


Answer:

p(x) = x3 – 3x2 + 5x – 3

g(x) = x2 – 2



On dividing them,

The quotient is x – 3.

And,

The remainder is 7x – 9.


Question 2.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 - x


Answer:

p(x) = x4 – 3x2 + 4x + 5

g(x) = x2 + 1 – x

On dividing them,

The quotient is x2 + x - 3

And,

The remainder is 8


Question 3.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

p(x) = x4 – 5x + 6, g(x) = 2 - x2


Answer:

p(x) = x4 – 5x + 6

g(x) = 2 – x2

On dividing them,



The quotient is –x2 – 2

And,

The remainder is -5x + 10


Question 4.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

t2 – 3, 2t4 + 3t3 – 2t2 – 9t - 12


Answer:

p(x) = 2t4 + 3t3 – 2t2 - 9t – 12

g(x) = t2 – 3


On dividing them,



The remainder is 0


Hence, yes t2 – 3 is a factor of 2t4 + 3t3 – 2t2 - 9t – 12


Question 5.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2


Answer:

p(x) = 3x4 + 5x3 – 7x2 + 2x + 2

g(x) = x2 + 3x + 1


On dividing them,



The remainder is 0

Hence, yes x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2


Question 6.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

x2 - 3x + 1, x5 - 4x3 + x2 + 3x + 1


Answer:

p(x) = x5 – 4x3 + x2 + 3x + 1

g(x) = x3 – 3x + 1

On dividing them,


The remainder is 2 ≠ 0

Hence, no x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1


Question 7.

Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5,

if two of its zeroes are  and 


Answer:

Two zeroes are  and 

is a factor


is a factor


is a factor


Now,



Therefore, 3x2 + 6x +3 is also a factor


Dividing 3x2 + 6x +3 by 3,


We get,


x2 + 2x +1


Factorising x2 + 2x +1,


x2 + 2x +1 = 0


⇒ x2 + x + x +1 = 0


⇒ x(x+ 1) + 1(x +1) = 0


⇒ (x+ 1)(x +1) = 0


⇒ x= -1, -1


Therefore, the other zeroes are -1 and -1.



Question 8.

On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and – 2x + 4 respectively. Find g(x)


Answer:

Dividend = x3 – 3x2 + x + 2

Quotient = x – 2


Remainder = -2x + 4


Dividend = Divisor × Quotient + Remainder


⇒ x3 – 3x2 + x + 2 = Divisor ×(x – 2) + (-2x + 4)


⇒ x3 – 3x2 + x + 2 – (-2x + 4) = Divisor ×(x – 2)


⇒ x3 – 3x2 + x + 2 + 2x – 4 = Divisor ×(x – 2)


⇒ x3 – 3x2 + 3x – 2 = Divisor ×(x – 2)



Therefore, g(x) = x2 – x + 1



Question 9.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg p (x) = deg q (x)


Answer:

Let g(x) = 2

And p(x) = 2x2 – 2x + 14


Then, dividing p(x) by g(x) gives.


q(x) = x2 – x + 7


and,


r(x) = 0


Deg(p(x)) = Deg(q(x)) = 2


Hence,


g(x) = 2


p(x) = 2x2 – 2x + 14


q(x) = x2 – x + 7


r(x) = 0



Question 10.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg q (x) = deg r (x)


Answer:

Let g(x) = x + 1

And p(x) = x2 + 3


Then, dividing p(x) by g(x) gives


q(x) = x


and,


r(x) = -x + 3


Deg(r(x)) = Deg(q(x)) = 1


Hence,


g(x) = x + 1


p(x) = x2 + 3


q(x) = x


r(x) = -x + 3



Question 11.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg r (x) = 0


Answer:

Let g(x) = 3

And p(x) = 3x + 3


Then, dividing p(x) by g(x) gives


q(x) = x + 1


and,


r(x) = 0


Deg(r(x)) = 0


Hence,


g(x) = 3


p(x) = 3x + 3


q(x) = x + 1


r(x) = 0


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