Practice Set 3.5 Polynomials Class 9th Mathematics Part I MHB Solution

Practice Set 3.5 Polynomials Class 9th Mathematics Part I MHB Solution

Practice Set 3.5
Question 1.

Find the value of polynomial using given values for x.

i. x = 3 ii. x = -1

iii. x=0


Answer:

The given polynomial is as follows:

2x – 2x3 + 7


(i) Put x = 3 in the given polynomial


= 2 × 3 – 2 × (3)3 + 7


= 6 – 2 × 27 + 7


= 6 -54 + 7


= -41


(ii) Put x = -1 in the given polynomial


= 2 × (-1) – 2 × (-1)3 + 7


= -2 + 2 + 7


= 7


(iii) Put x = 0 in the given polynomial


= 2 × 0 – 2 × (0)3 + 7


= 0 + 0 + 7


= 7



Question 2.

For each of the following polynomial, find and

p(x) = x3


Answer:

To find: p (1)

Put x = 1 in the given polynomial


p (1) = (1)3


p (1) = 1


To find: p (0)


Put x = 0 in the given polynomial


p (0) = 0


To find: p (-2)


Put x = -2 in the given polynomial


p (-2) = (-2)3


p (-2) = -8



Question 3.

For each of the following polynomial, find and

p(y) = y2 – 2y + 5


Answer:

To find: p (1)

Put x = 1 in the given polynomial


p (1) = (1)2 – 2 × 1 + 5


= 1 – 2 + 5


= -1 + 5


= 4


To find: p (0)


Put x = 0 in the given polynomial


p (0) = 0 – 0 + 5


= 5


To find: p (-2)


Put x = -2 in the given polynomial


p (-2) = (-2)2 – 2 × (-2) + 5


= 4 + 4 + 5


p (-2) = 13



Question 4.

For each of the following polynomial, find and

p(x) = x4 – 2x2 - x


Answer:

To find: p (1)

Put x = 1 in the given polynomial


p (1) = (1)4 – 2 × (1)2 – 1


= 1 – 2 – 1


p (1) = -2


To find: p (0)


Put x = 0 in the given polynomial


p (0) = 0 – 0 – 1


p (0) = -1


To find: p (-2)


Put x = -2 in the given polynomial


p (-2) = (-2)4 – 2 × (-2)2 – 1


= 16 – 2 × 4 – 1


= 16 -8 -1


= 7



Question 5.

If the value of the polynomial m3 + 2m + a is 12 for m =2, then find the value of a.


Answer:

Put m = 2 throughout the polynomial

(2)3 + 2 × 2 + a = 12


8 + 4 + a = 12


12 + a = 12


a = 0



Question 6.

For the polynomial mx2 + 2x + 3 if p(-1) = 7 then find m.


Answer:

Put x = -1 throughout the polynomial

m × (-1)2 – 2 × (-1) + 3 = 7


m + 2 + 3 = 7


m + 5 = 7


m = 7-5


m = 2



Question 7.

Divide the first polynomial by the second polynomial and find the remainder using factor theorem.

(x2 – 7x + 9); (x + 1)


Answer:

The coefficient form of the first polynomial is as follows:

⟹ 1.x2 – 7.x + 9


⟹ (1, -7, 9)



Therefore, the remainder is = 17



Question 8.

Divide the first polynomial by the second polynomial and find the remainder using factor theorem.

(2x3 – 2x2 + ax – a); (x-a)


Answer:

The coefficient form of the first polynomial is as follows:

⟹ 2.x3 – 2.x2 + a.x – a


(2, -2, a, -a)



The remainder is = 2a3-a2 –a



Question 9.

Divide the first polynomial by the second polynomial and find the remainder using factor theorem.

(54m3 + 18m2 – 27m + 5); (m-3)


Answer:

The coefficient form of the first polynomial is as follows:

⟹ 54.m3 + 18.m2 – 27.m + 5


(54, 18, -27, 5)



The last term in the synthetic division is the remainder of the given division method.


The remainder is =1544.



Question 10.

If the polynomial y3 – 5y2 + 7y + m is divided by y+2 and the remainder is 50 then find the value of m.


Answer:

Put y = -2 throughout the given polynomial

(-2)3 – 5 × (-2)2 + 7 × (-2) + m = 50


-8 – 20 -14 + m = 50


-28 + 14 + m = 50


-14 + m = 50


m = 50 + 14


m = 64



Question 11.

Use factor theorem to determine whether x+3 is factor of x2 + 2x – 3 or not.


Answer:

The coefficient form of the polynomial:

⟹ 1.x2 + 2.x – 3


(1, 2, -3)



Since the last term is zero which indicates that the polynomial on dividing by x + 3 leaves no remainder and hence x + 3 is the factor of x2 + 2x – 3



Question 12.

If (x-2) is a factor of x3 – mx2 + 10x - 20 then find the value of m.


Answer:

Put x = 2 throughout the polynomial and equate it to zero as x -2 is the factor of the given polynomial.

(2)3 – m × (2)2 + 10 × 2 – 20 = 0


8 – 4m + 20 – 20 = 0


4m = 8


m = 2


So for x-2 to be the factor of the given polynomial, the value of m = 2.



Question 13.

By using factor theorem in the following examples, determine whether q(x) is a factor p(x) or not.

p(x) = x3 – x2 – x – 1, q(x) = x - 1


Answer:

The coefficient form of p(x) is:

p(x) = 1.x3 – 1.x2 – 1.x – 1


(2, -1, 0, -45)



Since the last term is not term is not zero, q(x) is not the factor of p(x).



Question 14.

By using factor theorem in the following examples, determine whether q(x) is a factor p(x) or not.

p(x) = 2x3 – x2 – 45, q(x) = x - 3


Answer:

The coefficient form of p(x) is:

p(x) = 2.x3 – 1.x2 + 0.x – 45


(2, -1, 0, -45)



Since the last term is not term is not zero, q(x) is not the factor of p(x).



Question 15.

If (x31 + 31) is divided by (x+1) then find the remainder.


Answer:

Put x= -1 throughout the polynomial

x(-1) = (-1)31 + 31


= -1 + 31


= 30


So the remainder is 30.



Question 16.

Show that m-1 is a factor of m21 - 1 and m22 - 1


Answer:

Put m = 1 in the first polynomial

m (1) = (1)21 – 1


= 1 – 1


= 0


So m-1 is the factor of m21 -1


Put m = 1 in the second polynomial


m (1) = (1)22 – 1


= 1 – 1


= 0


So m-1 is the factor of m22 -1



Question 17.

If x-2 and x – 1/2 both are the factors of the polynomial nx2 – 5x + m, then show that m= n=2


Answer:

Put x = 2 in the given polynomial:

n× 4 – 5× 2 + m = 0


4n + m = 10 ……………………………. (i)


Put x = 1/2 in the given polynomial


n × 1/4 - 5 × 1/2 + m = 0


n + 4m – 10 = 0


n + 4m = 10 ……………………………….. (ii)


Since the RHS of both the equations is same,


4n + m = n + 4m


3n = 3m


n = m


Put n = m in the equation (i)


4m + m = 10


5m = 10


m = 2 = n



Question 18.

If p(x) = 2+5x then p(2) + p(-2)-p(1)


Answer:

Put x = 2 in the polynomial

p(2) = 2 + 5× 2


= 2 + 10


= 12


Put x = -2 in the polynomial


p(-2) = 2 + (5× -2)


= 2 – 10


= -8


Put x = 1 in the given polynomial


P(1) = 2 + 5× 1


= 2 + 5


= 7


p(2) + p(-2) – p(1) = 12 – 8 - 7


= -3



Question 19.

If p(x)=2x2-5√3+5 then p(5√3).


Answer:

Put x = 5√3 in the given polynomial

p(5√3) = 2×(5√3)2 – 5√3 × 5√3 + 5


= 2 × 25 × 3 – 25 × 3 + 5


= 150 – 75 + 5


p (5√3) = 80


PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!