# Exercise 2.5 - Chapter 2 Polynomials class 9 ncert solutions Maths PDF Download [2022-2023]

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Question 1:

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

(i) x3 + x2 + x + 1 (ii) x4 + x3 + x2 + x + 1

(iii) x4 + 3x3 + 3x2 + x + 1 (iv)

(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x3 + x2 + 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) = x4 + x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x4 + x3 + x2 + 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) = x4 + 3x3 + 3x2 + 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) = 2x3 + x2 − 2x − 1, g(x) = x + 1

(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2

(iii) p(x) = x3 − 4 x2 + x + 6, g(x) = x − 3

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

p(x) = 2x3 + x2 − 2x − 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) = x3 +3x2 + 3x + 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) = x3 − 4 x2 + 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) = x2 + x + (ii)

(iii)   (iv) p(x) = kx2 − 3x + k

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

(i) p(x) = x2 + 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) = kx2 − 3x + k

⇒ p(1) = 0

⇒ k(1)2 − 3(1) + k = 0

⇒ k − 3 + k = 0

⇒ 2− 3 = 0

Therefore, the value of k is .

Question 4:

Factorise:

(i) 12x2 − 7x + 1 (ii) 2x2 + 7x + 3

(iii) 6x2 + 5x − 6 (iv) 3x2 − x − 4

(i) 12x2 − 7x + 1

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

Here, 12x2 − 7x + 1 = 12x2 − 4− 3x + 1

= 4(3− 1) − 1 (3− 1)

= (3− 1) (4− 1)

(ii) 2x2 + 7x + 3

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

They are p = 6 and = 1.

Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3

= 2(+ 3) + 1 (+ 3)

= (x + 3) (2x+ 1)

(iii) 6x2 + 5x − 6

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

They are p = 9 and = −4.

Here,

6x2 + 5x − 6 = 6x2 + 9x − 4x − 6

= 3(2+ 3) − 2 (2+ 3)

= (2x + 3) (3− 2)

(iv) 3x2 − x − 4

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

and q = −1.

They are p = −4 and = 3.

Here,

3x2 − x − 4 = 3x2 − 4x + 3x − 4

(3− 4) + 1 (3− 4)

= (3x − 4) (+ 1)

Question 5:

Factorise:

(i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9− 5

(iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1

(i) Let p(x) = x3 − 2x2 − 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 x3 − 2x2 − x + 2 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ x3 − 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0

= (x + 1) [x2 − 2x − 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) = x3 − 3x2 − 9− 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 x3 + 3x2 − 9− 5 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ x3 − 3x2 − 9− 5 = (+ 1) (x2 − 4x − 5) + 0

= (+ 1) (x2 − 5x + x − 5)

(x + 1) [((x − 5) +1 (x − 5)]

= (x + 1) (x − 5) (x + 1)

= (x − 5) (x + 1) (x + 1)

(iii) Let p(x) = x3 + 13x2 + 32x + 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, + 1 is a factor of this polynomial p(x).

Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x3 + 13x2 + 32x + 20 = (+ 1) (x2 + 12x + 20) + 0

= (+ 1) (x2 + 10x + 2+ 20)

= (x + 1) [x (+ 10) + 2 (+ 10)]

= (x + 1) (+ 10) (+ 2)

= (x + 1) (x + 2) (x + 10)

(iv) Let p(y) = 2y3 + y2 − 2y − 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 2y3 + y2 − 2y − 1 by y ­− 1.

p(y) = 2y3 + y2 − 2y − 1

= (− 1) (2y2 +3y + 1)

= (− 1) (2y2 +2y + y +1)

= (− 1) [2(+ 1) + 1 (+ 1)]

= (− 1) (+ 1) (2+ 1)

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