Class 9 Maths

# Polynomials

## Exercise 2.4 Part 5

Question 5: Factorise:

(i) x^3 - 2x^2 - x + 2

Answer: p(x) = x^3 - 2x^2 - x + 2

Here constant term = 2
The prime factor of constant term 2 = 1 and 2
By trial method, if p(1) = 0, then, (x – 1) is one of the factor of the given polynomial.

Hence, p(1) = 1^3 - 2 xx 1^2 - 1 + 2
= 1 – 2 - 1 + 2
Or, p(1) = 0

Since, p(1) = 0, thus, (x - 1) is one of the factor of given polynomial.
Thus, given polynomial can be written as

x^3 - x^2 - x^2 + x – 2x + 2
= x^2(x - 1) – x( x – 1) – 2(x – 1)
(by taking (x – 1) as common)
= (x – 1)(x^2 - x – 2)

Note: second factor other than (x - 1) can be calculated by using long division.

Now, factorise the second factor, ie. (x^2 - x – 2)

Using splitting middle term method

= (x – 1)(x^2 + x – 2x – 2)
= (x – 1)[x(x + 1) – 2(x + 1)]
= (x – 1)(x + 1)(x – 2)
Hence, factor of given polynomial = (x – 1)(x + 1)(x – 2)

(ii) x^3 - 3x^2 - 9x – 5

Answer: Given, x^3 - 3x^2 - 9x – 5

Here, prime factors of constant term 5 = 1 and 5
Thus, if p(1) = 0, then (x - 1) will be one of the factor of the given polynomial
Or, if p(5) = 0, then ( x – 5) will be one of the factor of the given polynomial.

Hence, (1) = 1^3 - 3 xx 1^2 - 9 xx 1 – 5
= 1 – 3 – 9 – 5 = - 16

Since, p(1) ≠ 0, thus, ( x- 1) is not a factor of given polynomial.
Now,

p(5) = 5^3 - 3 xx 5^2 - 9 x 5 – 5
= 125 – 75 – 45 – 5 = 0

Since, p(5) = 0, thus, (x – 5) is one of the factor of given polynomial.
Now, given polynomial can be written as follows:

= x^3 - 5x^2 + 2x^2 - 10x + x – 5
= x^2(x – 5) + 2x(x – 5) + 1 (x – 5)
(by taking (x – 5) as common)
= (x – 5)(x^2 + 2x + 1)

Now, factorise the second factor, i.e. (x^2 + 2x + 1) by splitting the middle term.

= (x – 5)(x^2 + x + x + 1)
= (x – 5)[x(x + 1) + 1(x + 1)]
= (x – 5)(x+1)(x + 1)

(iii) x^3 + 13x^2 + 32x + 20

Answer: Given, p(x) = x^3 + 13x^2 + 32x + 20

Factors of 20 = (1 and 2), (2 and 10), (4 and 5)
Now,

p(1) = 1^3 + 13 xx 1^2 + 32 xx 1 + 20
= 1 + 13 + 32 + 20 = 66

Since, p(1) ≠ 0, thus, (x – 1) is not a factor of the given polynomial.
Now,

p( - 1) = (- 1)^3 + 13 xx( - 1)^2 + 32( - 1) + 20
= - 1 + 13 – 32 + 20 = 0

Since, p( -1) = 0, thus, (x + 1) is one of the factor of given polynomial.
Now, given polynomial can be written as:

x^3 + (x^2+ 12x^2 + (12x + 20x) + 20
= x^3 + x^2 + 12x^2 + 12x + 20x + 20
(by taking (x + 1) as common)
= x^2(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x^2 + 12x + 20)
= (x + 1)[x^2 + 2x + 10x + 20x]
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)

(iv) 2y^3 + y^2 - 2y – 1

Answer: Given, 2y^3 + y^2 - 2y – 1

Factors of ab = 2 = (1 and 2)
By trial method we will calculate p(1) and p(2)
Now, Let y = 1

p(1) = 2 xx 1^3 + 1^2 - 2 xx 1 – 1
= 2 + 1 – 2 – 1 = 0

Since, p(1) = 0 thus, (y - 1) is one of the factor of given polynomial.
Thus, the given polynomial can be written as:

2y^3 - 2y^2 + 3y^2 - 3y + y – 1
= 2y^2(y – 1) + 2y(y – 1) + 1(y – 1)
= (y – 1)(2y^2 + 3y + 1)
= (y -1)(2y^2 + 2y + y + 1
= (y – 1) [2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)