**Question** 1:

Determine which of the following polynomials has (*x* + 1) a factor:

(i) *x*^{3} + *x*^{2} + *x* + 1 (ii) *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1

(iii) *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1 (iv)

**Answer**:

(i) If (*x* + 1) is a factor of *p*(*x*) = *x*^{3} + *x*^{2} + *x* + 1, then *p* (−1) must be zero, otherwise (*x* + 1) is not a factor of *p*(*x*).

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

*p*(−1) = (−1)^{3} + (−1)^{2} + (−1) + 1

= − 1 + 1 − 1 + 1 = 0

Hence, *x* + 1 is a factor of this polynomial.

(ii) If (*x* + 1) is a factor of *p*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1, then *p* (−1) must be zero, otherwise (*x* + 1) is not a factor of *p*(*x*).

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

*p*(−1) = (−1)^{4} + (−1)^{3} + (−1)^{2} + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1

As *p*(− 1) ≠ 0,

Therefore, *x* + 1 is not a factor of this polynomial.

(iii) If (*x* + 1) is a factor of polynomial *p*(*x*) = *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1, then *p*(−1) must be 0, otherwise (*x* + 1) is not a factor of this polynomial.

*p*(−1) = (−1)^{4} + 3(−1)^{3} + 3(−1)^{2} + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1

As *p*(−1) ≠ 0,

Therefore, *x* + 1 is not a factor of this polynomial.

(iv) If(*x* + 1) is a factor of polynomial *p*(*x*) = , then *p*(−1) must be 0, otherwise (*x* + 1) is not a factor of this polynomial.

As *p*(−1) ≠ 0,

Therefore, (*x* + 1) is not a factor of this polynomial.

**Question** 2:

Use the Factor Theorem to determine whether *g*(*x*) is a factor of *p*(*x*) in each of the following cases:

(i) *p*(*x*) = 2*x*^{3} + *x*^{2} − 2*x* − 1, *g*(*x*) = *x* + 1

(ii) *p*(*x*) = *x*^{3} + 3*x*^{2} + 3*x* + 1, *g*(*x*) = *x* + 2

(iii) *p*(*x*) = *x*^{3} − 4 *x*^{2} + *x* + 6, *g*(*x*) = *x* − 3

**Answer**:

(i) If *g*(*x*) = *x* + 1 is a factor of the given polynomial *p*(*x*), then *p*(−1) must be zero.

*p*(*x*) = 2*x*^{3} + *x*^{2} − 2*x* − 1

*p*(−1) = 2(−1)^{3} + (−1)^{2} − 2(−1) − 1

= 2(−1) + 1 + 2 − 1 = 0

Hence, *g*(*x*) = *x* + 1 is a factor of the given polynomial.

(ii) If *g*(*x*) = *x* + 2 is a factor of the given polynomial *p*(*x*), then *p*(−2) must

be 0.

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

*p*(−2) = (−2)^{3} + 3(−2)^{2} + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As *p*(−2) ≠ 0,

Hence, *g*(*x*) = *x* + 2 is not a factor of the given polynomial.

(iii) If *g*(*x*) = *x* − 3 is a factor of the given polynomial *p*(*x*), then *p*(3) must

be 0.

*p*(*x*) = *x*^{3} − 4 *x*^{2} + *x* + 6

*p*(3) = (3)^{3} − 4(3)^{2} + 3 + 6

= 27 − 36 + 9 = 0

Hence, *g*(*x*) = *x* − 3 is a factor of the given polynomial.

**Question** 3:

Find the value of *k*, if *x* − 1 is a factor of *p*(*x*) in each of the following cases:

(i) *p*(*x*) = *x*^{2} + *x* + *k *(ii)* ** *

(iii) * *(iv) *p*(*x*) = *kx*^{2} − 3*x* + *k*

**Answer**:

If *x* − 1 is a factor of polynomial *p*(*x*), then *p*(1) must be 0.

(i) *p*(*x*) = *x*^{2} + *x* + *k*

*p*(1) = 0

⇒ (1)^{2} + 1 + *k* = 0

⇒ 2 + *k* = 0 ⇒ *k* = −2

Therefore, the value of* k *is −2*.*

(ii)* ** *

*p*(1) = 0

(iii)* ** *

*p*(1) = 0

(iv) *p*(*x*) = *kx*^{2} − 3*x* + *k*

⇒ *p*(1) = 0

⇒ *k*(1)^{2} − 3(1) + *k* = 0

⇒ *k* − 3 + *k* = 0

⇒ 2*k *− 3 = 0

Therefore, the value of* k *is .

**Question** 4:

Factorise:

(i) 12*x*^{2} − 7*x* + 1 (ii) 2*x*^{2} + 7*x* + 3

(iii) 6*x*^{2} + 5*x* − 6 (iv) 3*x*^{2} − *x* − 4

**Answer**:

(i) 12*x*^{2} − 7*x* + 1

We can find two numbers such that *pq* = 12 × 1 = 12 and *p *+ *q* = −7. They are *p* = −4 and *q *= −3.

Here, 12*x*^{2} − 7*x* + 1 = 12*x*^{2} − 4*x *− 3*x* + 1

= 4*x *(3*x *− 1) − 1 (3*x *− 1)

= (3*x *− 1) (4*x *− 1)

(ii) 2*x*^{2} + 7*x* + 3

We can find two numbers such that *pq* = 2 × 3 = 6 and *p *+ *q* = 7.

They are *p* = 6 and *q *= 1.

Here, 2*x*^{2} + 7*x* + 3 = 2*x*^{2} + 6*x* + *x* + 3

= 2*x *(*x *+ 3) + 1 (*x *+ 3)

= (*x* + 3) (2*x+ *1)

(iii) 6*x*^{2} + 5*x* − 6

We can find two numbers such that *pq* = −36 and *p *+ *q* = 5.

They are *p* = 9 and *q *= −4.

Here,

6*x*^{2} + 5*x* − 6 = 6*x*^{2} + 9*x* − 4*x* − 6

= 3*x *(2*x *+ 3) − 2 (2*x *+ 3)

= (2*x* + 3) (3*x *− 2)

(iv) 3*x*^{2} − *x* − 4

We can find two numbers such that *pq* = 3 × (− 4) = −12

and *p *+ *q* = −1.

They are *p* = −4 and *q *= 3.

Here,

3*x*^{2} − *x* − 4 = 3*x*^{2} − 4*x* + 3*x* − 4

= *x *(3*x *− 4) + 1 (3*x *− 4)

= (3*x* − 4) (*x *+ 1)

**Question** 5:

Factorise:

(i) *x*^{3} − 2*x*^{2} − *x* + 2 (ii) *x*^{3} + 3*x*^{2} −9*x *− 5

(iii) *x*^{3} + 13*x*^{2} + 32*x* + 20 (iv) 2*y*^{3} + *y*^{2} − 2*y* − 1

**Answer**:

(i) Let *p*(*x*) = *x*^{3} − 2*x*^{2} − *x* + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

*p*(−1) = (−1)^{3} − 2(−1)^{2} − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (*x* +1 ) is factor of polynomial *p*(*x*).

Let us find the quotient on dividing *x*^{3} − 2*x*^{2} − *x* + 2 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 2*x*^{2} − *x* + 2 = (*x* + 1) (*x*^{2} − 3*x* + 2) + 0

= (*x* + 1) [*x*^{2} − 2*x* − *x* + 2]

= (*x* + 1) [*x* (*x* − 2) − 1 (*x* − 2)]

= (*x* + 1) (*x* − 1) (*x* − 2)

= (*x* − 2) (*x* − 1) (*x* + 1)

(ii) Let *p*(*x*) = *x*^{3} − 3*x*^{2} − 9*x *− 5

All the factors of 5 have to be considered. These are ±1, ± 5.

By trial method,

*p*(−1) = (−1)^{3} − 3(−1)^{2} − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, *x* + 1 is a factor of this polynomial.

Let us find the quotient on dividing *x*^{3} + 3*x*^{2} − 9*x *− 5 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 3*x*^{2} − 9*x *− 5 = (*x *+ 1) (*x*^{2} − 4*x* − 5) + 0

= (*x *+ 1) (*x*^{2} − 5*x* + *x* − 5)

= *(x* + 1) [(*x *(*x* − 5) +1 (*x* − 5)]

= (*x* + 1) (*x* − 5) (*x* + 1)

= (*x* − 5) (*x* + 1) (*x* + 1)

(iii) Let *p*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20

All the factors of 20 have to be considered. Some of them are ±1,

± 2, ± 4, ± 5 ……

By trial method,

*p*(−1) = (−1)^{3} + 13(−1)^{2} + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0

As *p*(−1) is zero, therefore, *x *+ 1 is a factor of this polynomial *p*(*x*).

Let us find the quotient on dividing *x*^{3} + 13*x*^{2} + 32*x* + 20 by (*x* + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

*x*^{3} + 13*x*^{2} + 32*x* + 20 = (*x *+ 1) (*x*^{2} + 12*x* + 20) + 0

= (*x *+ 1) (*x*^{2} + 10*x* + 2*x *+ 20)

= (*x* + 1) [*x* (*x *+ 10) + 2 (*x *+ 10)]

= (*x* + 1) (*x *+ 10) (*x *+ 2)

= (*x* + 1) (*x* + 2) (*x* + 10)

(iv) Let *p*(*y*) = 2*y*^{3} + *y*^{2} − 2*y* − 1

By trial method,

*p*(1) = 2 ( 1)^{3} + (1)^{2} − 2( 1) − 1

= 2 + 1 − 2 − 1= 0

Therefore, *y* − 1 is a factor of this polynomial.

Let us find the quotient on dividing 2*y*^{3} + *y*^{2} − 2*y* − 1 by *y* − 1.

*p*(*y*) = 2*y*^{3} + *y*^{2} − 2*y* − 1

= (*y *− 1) (2*y*^{2} +3y + 1)

= (*y *− 1) (2*y*^{2} +2y + *y* +1)

= (*y *− 1) [2*y *(*y *+ 1) + 1 (*y *+ 1)]

= (*y *− 1) (*y *+ 1) (2*y *+ 1)

We hope we have given the updated helpful content for our users of Exercise 2.5 - Chapter 2 Polynomials class 9 ncert solutions Maths
in an easily accessible format to help you in preparing adequately. You can also download and practise these - Exercise 2.5 - Chapter 2 Polynomials class 9 ncert solutions Maths in PDF For Free Download 2022 - 2023 to get a thorough knowledge of the concepts
and also Search and Check Out Our Latest Study Materials :

All Books Answer Keys Solutions and Guides
From Class 3 to 12 and NCERT Solutions | NCERT Books| RD Sharma Solutions | NCERT Exemplar Problems | CBSE Sample Papers in detail Step by step | Samacheer Kalvi Books Tamilnadu State Board Text Books
Solutions for Class 12th, 11th, 10th, 9th, 8th, 7th, and 6th Standards.
We do offer free support materials, 10th maths guide, 10th science guide, 10th tamil guide, 10th social guide, 10th english guide, science 10th guide
and also for all the classes books and guide to all the students who sign up for SaraNextGen. Apart from traditional textbook queries, we've got conjointly provided further high order level thinking issues,
that are seemingly to be expected in boards and competitive exams. It includes conceptual queries, MCQs, Long and short answer type queries, etc.
These Questions are designed to learn each student and academics by providing chapter-wise further issues focusing totally on testing conceptual information with applications.
Here, we've got provided a number of the necessary ways in which within which the solutions of all the Questions will profit students of class 3 to 12. if you want generate the
Tests Series in pdf for free you can instantly generate it at TEST GENERATOR - CBSE JEE NEET CUET UPSC | Self Study App For Teachers, Students & Parents. Download it in PDF
and also you can Generate 100 Questions For Free. If you have any queries, drop a message to us and we will get back to you at the earliest.