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## Kerala State Syllabus 10th Standard Maths Solutions Chapter 10 Polynomials

### Polynomials Textbook Questions & Answers

Textbook Page No. 237

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**Polynomials Class 10 Kerala Syllabus Questions 1.**

Write the second degree polynomials. given below as the product of two first degree polynomials. Find also the solutions of the equation p(x) = 0 in each.

i. p(x) = x^{2} – 7x+12

ii. p(x) = x^{2} + 7x + 12

iii.p (x) = x^{2} – 8x +12

iv. p(x) = x^{2} + 13x +12

v. p (x) = x^{2} + 12x – 13

vi. p (x) = x^{2} – 12x – 13

Answer:

i. p (x) = x^{2} – 7x + 12

a + b = –7, ab = 12

a = –3, b = –4

x^{2} – 7x + 12 = (x – 3) (x – 4)

x^{2} – 7x + 12 = 0

(x – 3) (x – 4) = 0

x – 3 = 0, x – 4 = 0

x = 3, x = 4

ii. p(x) = x^{2} + 7x + 12

a + b = 7, ab= 12 a = 3, b = 4

x^{2} + 7x + 12 = (x + 3) (x + 4)

x^{2} + 7x + 12 = 0

(x + 3)(x + 4) = 0

x = 3, x = 4

iii. p(x) = x^{2 –} 8x + 12

a + b = 8, ab= 12

a = 6, b = –2

x^{2} – 8x + 12 = (x – 6) (x – 2)

(x – 6) (x – 2) = 0

x = 6, x = 2

iv. p(x) =x^{2} + 13x + 12

a+b = 13, ab = 12

a =12, b=1

(x^{2} +13x + 12) = (x + 12) (x + 1)

x^{2}+13x + 12 = 0

(x+ 12) (x+ 1) = 0

x+ 12 = 0, x+ 1 = 0

x = -12 ,x = –1

v. p(x) = x^{2 }+ 12x – 13

a = –13, b = 1

x^{2 }+ 12x – 13 = (x + 13)(x – 1)

x + 13 = 0, x – 1 =0

x = –13, x= 1 .

vi. p(x) = x^{2} – 12x – 13

x^{2} – 12x – 13 = (x – a) (x – b)

= x^{2} – (a + b) x + ab

a + b= 12 ab = –13

(a – b)^{2} =(a + b)^{2} – 4ab

= (12)^{2} – 4x – 13 = 196

a – b = 14

a + b= 12

a= 13; b = –1

x^{2} – 12x – 13 = (x – 13)(x + 1)

x – 13 = 0, x + 1 =0

x= 13 ,x = –1

Textbook Page No. 240

Is this a Polynomial Calculator is an online tool that helps to calculate the result of addition, subtraction, multiplication, and division of two polynomials.

**Sslc Maths Chapter 10 Kerala Syllabus Questions 1.**

In each pair of polynomials given below, find the number to be subtracted from the first to get a polynomial for which the second is as factor. Find also the second factor of the polynomial got on subtracting the number.

i. x^{2} – 3x + 5, x – 4

ii. x^{2} – 3x + 5, x + 4

iii. x^{2} + 5x – 7, x – 1

iv. x^{2} – 4x – 3, x – 1

Answer:

p(x) = x^{2} – 3x + 5

If x – 4 is a factor, p(4) = 0

p(4) = (4)^{2} – 3 x 4 + 5 = 9

For x – 4 to become a factor of p (4) must be equal to zero.

For p (4) = 0 here we have to subtract 9 from p(x).

That is, add -9 to p(x) for (x – 4) become a factor.

∴ p(x) = x^{2} – 3x + 5 – 9 = x^{2} – 3x – 4

x^{2} – 3x – 4 = (x – a) (x – b)

= x^{2} – (a + b) x + ab

a + b = 3

ab = –4

(a – b)^{2} = (a + b)^{2} – 4 ab

= (3)^{2} – 4x – 4 = 25

a – b = 5

a + b = 3

a = 4; b = –1

x^{2} – 3x – 4 = (x – 4)(x + 1)

Second factor is (x +1)

ii. p(x) = x^{2} – 3x + 5

If x + 4 is a factor, p(–4) = 0

p(–4) = (–4)^{2} – 3x – 4 + 5 = 33

For x + 4 to become a factor of p (–4)

must be equal to zero.

For p (–4) = 0 here we have to subtract 33 from p(x).

That is, add –33 to p(x) for (x + 4) become a factor.

∴p(x) = x^{2} – 3x + 5 – 33 = x^{2} – 3x – 28

x^{2} – 3x – 28 = (x – a) (x – b)

= x^{2} – (a+b) x + ab

a + b = 3

ab = -28

(a – b)^{2} = (a + b)^{2} – 4ab

= (3)^{2} – 4x – 28 = 121

a – b= 11

a + b = 3

a = 7;

b = -4 x^{2} – 3x – 28 = (x – 7)(x + 4)

Second factor is (x – 7)

iii. p(x) = x^{2} + 5x – 7

If x – 1 is a factor, p(1) = 0

p(1) = (1)^{2} +5 x 1 – 7 = –1

For x – 1 to become a factor of p (1) must be equal to zero.

For p (1) = 0 here we have to subtract –1 from p(x).

That is, add 1 to p(x) for (x – 1) become a factor.

p(x) = x^{2} + 5x – 7 + 1 = x^{2} + 5x – 6

x^{2} + 5x – 6 = (x – a) (x – b)

a + b = 5

ab = –6

a = –6; b= 1

x^{2} + 5x – 6 = (x + 6)(x – 1)

Second factor is (x + 6)

iv. p(x) = x^{2} – 4x – 3

If x – 1 is a factor, p(1) = 0

p(1) = (1)^{2} – 4 x 1 – 3 = –6

For x – 1 to become a factor of p (1) must

be equal to zero.

For p (1) = 0 here we have to subtract -6 from p(x).

That is, add 6 to p(x) for (x – 1) become a factor.

∴ p(x) = x^{2} – 4 x – 3 + 6 = x^{2} – 4x +3

x^{2} – 4x + 3 = (x – a) (x – b)

a + b = –4 ab = 3

a =1; b = 3

x^{2} – 4x +3 = (x – 1)(x – 3)

Second factor is (x – 3)

Still, using a computation tool like a factor complex polynomials calculator, you can streamline things for you.

**Polynomials Class 10 State Syllabus Questions 2.**

In the polynomial x^{2} + kx + 6, what number must be taken ask to get a polynomial for which x –1 is a factor? Find also the other factor of that polynomial.

Answer:

p (x) = x^{2} + kx + 6

If (x – 1) is a factor of p(x)

then p(1) = 0

p(1) = 12 + k x 1 + 6 = 7 + k

7 + k = 0

k= –7

∴ p(x) = x^{2} – 7x + 6

a + b = 7

ab = 6

a= 1, b = 6

factors are (x – 1 )(x – 6)

Second factor is (x – 6)

**Polynomials Class 10 Kerala Syllabus Questions 3.**

In the polynomial kx^{2} + 2x – 5, what number must be taken ask to get a polynomial for which x –1 is a factor?

Answer:

p(x) = kx^{2} + 2x – 5

If (x – 1) is a factor, then p(1) = 0

p(1) = k(l)^{2} + 2 x 1 – 5

= k + 2 – 5 = k – 3

k – 3 = 0 k = 3

Textbook Page No. 242

**Sslc Polynomials Questions And Answers Kerala Syllabus Question 1.**

Write the second degree polynomials given below as die product of two first degree polynomials:

i. x^{2} – 20x + 91

ii. x^{2} – 20x + 51

iii. x^{2} + 5x – 84

iv. 4x^{2} – 16x +15

v. x^{2} – x – 1

Answer:

p(x) = x^{2} – 20x + 91

We must solve the equation p(x) = 0

x^{2} – 20x + 91 = 0

**Sslc Maths Polynomials Solutions Kerala Syllabus Question 2.**

Prove that none of the polynomials below can be factored into a product of first degree polynomials:

i. x^{2} + x + l

ii. x^{2} – x + l

iii. x^{2} + 2x + 2

iv. x^{2} + 4x + 5

Answer:

i. x^{2} + x+1

We must solve the equation p(x) = 0

x^{2} + x +1 =0

no solutions

p(x) doesn’t have any first degree factors.

ii. x^{2} – x + 1

We must solve the equation p(x) = 0

x^{2} – x + 1 = 0

no solutions

p(x) doesn’t have any first degree factors

iii. x^{2} + 2x + 2

We must solve the equation p(x) = 0

x^{2} + 2x + 2 = 0

no solutions

p(x) doesn’t have any first degree factors.

iv. x^{2} + 4x + 5

We must solve the equation p(x) = 0

x^{2} + 4x + 5 = 0

no solutions

p(x) doesn’t have any first degree factors.

**Sslc Maths Chapter 10 Solutions Kerala Syllabus Question 3.**

In the polynomial p(x) = x^{2} + 4x + k, up to what number can we take ask, so that p(x) can be factorized as a product of two first degree polynomials?

Answer:

p(x) = x^{2} + 4x + k

p(x) = 0

x^{2} + 4x + k = 0

p(x) can be factorized as a product of two first degree polynomials, the \(\sqrt{16-4 k}>0\) So, k can take value up to 4.

### Polynomials Orukkam Questions & Answers

Worksheet 1

**Polynomials Class 10 Hsslive Kerala Syllabus Question 1.**

Write the product (x – 1) x (x + 1)

Find the product of (x – 1),(x + 1),(x + 2) If the poduct is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.

Answer:

(x – 1)x (x + 1) = x^{2} – 1

(x – 1) (x + 1) (x + 2)=(x^{2} – 1) (x + 2)

= x^{3} – x + 2x^{2 }– 2 = x^{3} + 2x^{2} – x – 2 = 0

p(x) = x^{3} + 2x^{2} – x – 2 = 0

p(1) =1 + 2 – 1 – 2 = 0

p(–1) = – 1 + 2 + 1 – 2 = 0

p(–2) = – 8 + 8 + 2 – 2 = 0

1, – 1, – 2 are the solutions of the equation p(x) = 0.

**Hss Live Guru 10th Maths Kerala Syllabus Question 2.**

Expand (x – a)(x – b). If x^{2} – 7x + 12 = (x – a)(x – b) then find a+b.

Also find ab Calculate the values of a, b. Write the factors of (x^{2} – 7x +12 ).

Find the solutions of (x^{2} – 7x + 12).

Answer:

(x – a) (x – b) = x^{2} – bx – ax + ab

= x^{2} – x (a + b) + ab

= x^{2} – 7x + 12 = (x – a)(x – b)

a + b = 7

ab = 12

(a + b)^{2} – 4ab = 7^{2} – 4 x 12 = 49 – 48 = 1 = a – b

a + b = 7

a – b = 1, 2a = 8 a = \(\frac { 8 }{ 2 }\) =4

b = 7 – 4 = 3 a = 4 and b = 3.

Solution of x^{2} – 7x + 12 is x^{2} – 7x + 12 = (x – 4) (x – 3)

Solutions = 4, 3

**Maths Questions And Answers For Class 10 Kerala Syllabus Question 3.**

If p(x) = x3 – 6×2 + 11x – 1 then find p(1), p(2), p(3). Find p(x) – p(1), p(x) – p(2), p(x) – p(3), p(x) – p(1). Write the solutions of p(x) – p(1) = 0

Polynomials Class 10 Worksheet with Answer:

p(x) = x^{3} – 6x^{2} + 11x – 1

p(1) = 1 – 6 + 11 – 1 = 5

p(2) = 8 – 24 + 22 – 1 = 5

p(3) = 27 – 54 + 33 – 1 = 5

p(x) – p(1) = x^{3} – 6x^{2} + 11x – 6

p(x) – p(2) = x^{3} – 6x^{2} + 11x – 6

p(x) – p(3) = x^{3} – 6x^{2} + 11x – 6

p(x) – p(1) = x^{3} – 6x^{2} + 11x – 6 = 0

If x = 1, 2, 3 then p(x) – p(1) = 0.

Factors of the equations are (x – 1), (x – 2), (x – 3).

Solutions of the equations are 1, 2, 3.

**Hss Live Maths 10th Kerala Syllabus Question 4.**

When p(x)is divided by (ax + b), the quotient is q(x)and the remainder is c. p(x) = (ax + b) x q(x) + c

When does the value of p(x) equal to c \(p\left(\frac{-b}{a}\right)=\left(a \times \frac{-b}{a}+b\right) \times q\left(\frac{-b}{a}\right)+c\)

What is the remainder when p(x)is divided by ax + b. When does(ax + b)becomethe factor of p(x).

Answer:

The remainder obtained when p(x) is divided by (ax + b) = \(p\left(\frac{-b}{a}\right)\)

When \(p\left(\frac{-b}{a}\right)\) = 0, then (ax + b) will be a factor of p(x).

Worksheet 2

**Hsslive Maths 10th Kerala Syllabus Question 5.**

Write the following as the product of first degree polynomials

1. x^{2} + 7x + 12

2. x^{2} + 3x + 2

3. x^{2} – 9x – 22

4. 2x^{2} + 5x – 3

Answer:

1. x^{2} + 7x + 12 = (x + 4)(x + 3)

2. x^{2} + 3x + 2 = (x + 2)(x + 1)

3. x^{2} – 9x – 22 =(x – 11)(x + 2)

4.2x^{2} + 5x – 3 =\(\left(x-\frac{1}{2}\right)\)(x+3)=(2x-l)(x+3)

**Hss Live Maths 10 Kerala Syllabus Question 6.**

Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.

Answer:

If p( 1) = 0, then x – 1 is a factor.

If p(–2) = 0, then x + 2 is a factor.

If p(2) = 0, then x – 2 is a factor.

p(x) = (x – 1) (x + 2) (x – 2) = x^{3} – x^{2} – 4x + 4

**Class 10 Kerala Syllabus Maths Solutions Question 7.**

Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)

Answer:

**Hsslive 10th Maths Kerala Syllabus Question 8.**

Prove that x^{2} + 2x +2 cannot be written as the product of first degree polynomials

Answer:

b2 – 4ac = 22 – 4 x 1 x 2 = –4 < 0

∴ x^{2} + 2x + 1 cannot be written as the product of first degree polynomials

Worksheet 3

**Polynomials Class 10 In Malayalam Kerala Syllabus Question 9.**

Find the remainder and quotient obtained by dividing x^{3} – 5x^{2} + 7x + 3by (x + 2).

Answer:

Let quotient = x^{2} + ax + b and remainder be c, then

x^{3} – 5x^{2} + 7x + 3 =(x + 2)(x^{2} + ax + b) + c +

= x^{3} – 5x^{2} + 7x + 3 = x^{3} + ax^{2} + bx + 2x^{2} + 2ax + 2b + c

= x^{3} +( a + 2) x^{2} + (b + 2a)x + 2b + c From this equation,

a + 2 = –5,

b + 2a = 7, 2b + c = 3

i.e., a = –7

b= 21

c = -39

quotient = x^{2} – 7x + 21,

remainder = –39

**Maths Polynomials Class 10 State Syllabus Question 10.**

Given x – 1 is a factor of x^{2} + ax + b. Prove that (a + b = –1)

Answer:

Let (x – 1) be a factor, then p(1)=0

p(1) = 12 + a x 1 + b = 0

i.e., a + b = –1

**Hss Guru Maths 10 Kerala Syllabus Question 11.**

p(x) = (4x^{2} – 1)(x + 2)Write p(x) as the product of first degree factors.Write p(x) in the form of a trird degree polynomial What is the remainder obtained by dhpding 4x^{3} + 6x^{2} – x + 2by (x + 2) .What is the re¬mainder obtained by dividing 4x^{3} + 6x^{2} – x + 1 by (2x – 1).

Answer:

### Polynomials SCERT Questions & Answers

Question 12.

Write the second-degree polynomial p(x)= x^{2} + x – 6 as the product of first-degree polynomials. Find also the solution of the equation p (x)=0 [Score: 4, Time: 7 minute]

Answer:

Question 13.

For what values of x, the polynomial 2x^{2} – 7x – 15 is equal to zero? Write this polynomial as the product of two first degree polynomials. [Score: 4, Time: 7 minute]

Answer:

Question 14.

Write the polynomial p(x) = x^{2} + 4x + 1 as the product of two first degree polynomials. Find the solution of the equation p (x) = 0. [Score: 4, Time: 5 minute]

Answer:

x^{2} + 4x + 1 = (x – a) (x – b) = x^{2} – (a + b) x + ab(1)

a + b = –4, ab = 1, a – b = 2√3

a = –2 + √3, b = –2 – √3 (1)

x^{2} + 4x + 1 =(x + 2+ √3 ) (x + 2 – √3 ) (1)

x^{2} + 4x + 1 = 0 =>(x + 2 + √3 )(x + 2 –√3 ) = 0 (1)

x = –2 – √3 ,or x = –2 + √3

Question 15.

In the polynomial p (x) = x^{2}+ ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b. [ Score: 4, Time: 5 minute]

Answer:

p (x) = x^{2} + ax + b

p(3 + √2) = 0, (x – 3 – √2) is a factor (1)

p(3 – √2 ) = 0, ( x – 3 + √2) is a factor (1)

p(x) = x^{2} + ax + b = (x – 3 – √2) (x – 3 + √2)

= (x – 3)^{2} – ( √2)^{2} (1)

x^{2} + ax + b = x^{2} – 6x + 7 (1)

Question 16.

What number should be added to the polynomial p(x) = x^{2} + x – 1, so that (x – 2)is a factor of the new polynomial. [Score: 4, Time: 6 minute]

Answer:

p (x) = x^{2} + x – 1, remainder p(2) (1)

p (2) = (2)^{2} + 2 – 1 = 5 (1)

For x – 2 to become a factor of p (2) must be equal to zero.

For p (2) = 0 here we have to substract 5 ffomp(x). (1)

That is, add –5 to p(x) for (x – 2) become a factor. (1)

Question 17.

What is the smallest natural number k, for which the polynomial 2x^{2} + kx + 6 can be written as a product of two first degree polynomials? Write down the polynomial using k and express it as the product of two first degree polynomials [Score: 4, Time: 8 minutes]

Answer:

Question 18.

Method to check whether(x – a), and (x + a)are factors of a polynomial P(x).

Check whether (x + 2) and (x – 5) are factors of the polynomial p(x) = x^{2} + 7x + 10 [Score: 4, Time: 6 minute]

Answer:

When a polynomial p(x) is divided by (x – a), if p(a) = 0 then (x – a) is a factor of p(x). When a polynomial p(x) is divided by (x + a), if p(–a) = 0 then (x + a) is a factor of p(x).

p(x) = x^{2} + 7x + 10

p(–2) = 4 – 14 + 10 = 0 (1)

∴ x + 2 is a factor (1)

Remainder p(5) = (5)^{2} + 7(5) +10 (1)

= 25 + 35 + 10 ≠ 0 (1)

∴ x – 5 is not a factor

Question 19.

When dividing x^{2} + ax + b by (x – 2) and (x – 3) the remainder is zero. What are the numbers a and b. [Score: 3, Time: 5 minute]

Answer:

p(x) = x^{2} + ax + b = (x – 3)(x – 2)

= x^{2} – 5x + 6

a = –5, b = 6 (3)

### Polynomials Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 20.

Check whether x – 1 is a factor of 3x^{3} – 2x^{2} – 3x + 2.

Answer:

P(1)=3 x 13 – 2 x 12 – 3 x 1 + 2 = 0

Therefore x – 1 is a factor.

Question 21.

If (x – 1) is to be a factor of p(x) = a^{2}x^{2} – 4ax + 4a – 1. What should be the value of ‘a’ ?

Answer:

p(1) = 0

a2 – 4a + 4a – 1 = 0

a2 – 1 =0

a = +1 or –1

Question 22.

Answer:

Question 23.

Prove that (x – 1) is a factor of x^{13} – 1

Answer:

p(x) = x^{13} – 1

p(1)= 1^{13} – 1 = 1 – 1=0

p(1) = 0

(x – 1) is a factor of p(x)

Question 24.

Write the solution of polynomials p(x) = x^{2} – 7x + 12.

Answer:

x^{2} – 7x + 12 = (x – 4)(x – 3)

p(x) = 0 (x – 4) ( x – 3) = 0

x = 4, x = 3

Question 25.

The quotient is x2 – 5x + 6 when the polynomial p(x) is divided by (x – 1), and the remainder is 7. Then

a. p(x) = (………. ) ( ……… ) + 7. Complete it

b. Find P(2).

Answer:

a. p(x) = (x^{2} – 5x + 6) (x – 1) + 7

b. p(2) = 4 – 10 + 6 + 7 = 7

Short Answer Type Questions (Score 3)

Question 26.

a. Find the remainder when

x^{3} – 4x^{3} + 12x – 45 is divided by (x – 2)?

b. Find the value of k if the remainder is zero on dividing 2x^{3} + 4x^{2} – 10x + k by (x – 1)

Answer:

a. p(x) = x^{3} – 4x^{2} – 12x – 45

remainder = p(2) = 23 – 4(2)^{2} + 12 x 2 – 45 = 32 – 61 = –29

b. p(x) = 2x^{3} + 4x^{2} – 10x + k; p(l) = 0

=> 2 x 13 + 4 x 12 – 10 x 1 + k = 0

2 + 4 – 10 + k = 0 – 4 + k = 0; k = 4

Question 27.

Factorise 3x^{2} + 5x + 2 completely.

Answer:

Question 28.

Which first-degree polynomial is added to the polynomial 5x^{3} + 3x^{2} to get x^{2} – 1 as a factor.

Answer:

p(x) = 5x^{3} +3x^{2} + ax + b

∴ x^{2} – 1 isafoctorthen p(1), p(–1) will be zero.

p(1) = 5 x 1 + 3 x 1 + a x 1 + b = 0

= 5x – 1 + 3 x 1 + ax – 1 + b = 0

a + b = –8 (1)

= 5x – 1 + 3 x 1 + ax – 1 + b = 0

–a + b = 2 (2)

Find the solutions of the equation

b = –3, a = –5 added polynomial = –5x – 3

Long Answer Type Questions (Score 4)

Question 29.

a. Find the value of k if remainder when 5x^{3} + 4x – 11x + k is divided by (x – 1) is 0.

b. When x^{3} – 2x^{2} + kx + 7 is divided by (x – 4) remainder is 11. Find k.

Answer:

Question 30.

a Show that the polynomial x + x + 1 has no first degree factors,

b. What is the remainder when the polynomial (x – 1) (x – 2) (x – 3) is divided by (x – 1)?

c. When (x – 1) (x – 2) (x – 3) + 2x + k is divided by (x – 1), the remainder is 10.

Then find out the remainder when it is divided by (x – 2).

Answer:

Long Answer Type Questions (Score 5)

Question 31.

Write 2x^{2} + 5x + 3 as a product of two first degree polynomials.

Answer:

Question 32.

If(x + 1)and(x – 1) are factors of x3 + 2x^{2} + px +q, find p and q.

Answer:

.