Class 9 Maths

# Polynomials

## Exercise 2.5 Part 1

Question 1: Use suitable identities to find the following products:

(i) (x + 4)(x + 10)

Answer: Given, (x + 4)(x + 10)
We know that, (x + a)(x + b) = x^2 + (a + b) x + ab
Here, a = 4 and b = 10
Therefore, (x + 4)(x + 10) = x^2 + (4 + 10) x + 4 xx 10
= x^2 + 14x + 40 Answer

(ii) (x + 8)(x – 10)

Answer: Given, (x + 8)(x – 10)
= (x + 8)(x + (–10))
Here, a = 8 and b = – 10
Using identity (x + a)(x + b) = x^2 + (a + b) x + ab
given expression can be written as
x^2 + (8 +(– 10))x + 8 xx (– 10)
= x^2 +(8 – 10)x – 80
= x^2 + 2x – 80 Answer

(iii) (3x + 4)(3x – 5)

Answer: Given, (3x + 4)(3x – 5)
Here, x = 3x, a = 4 and b = - 5
Therefore, using identity (x + a)(x + b) = x^2 + (a + b) x + ab
given expression can be written as
(3x)^2 + (4 + (-5)) xx 3x + (4 (- 5))
= 9x^2 + ( 4 – 5)× 3x + (- 20)
= 9x^2 + ( - 1) xx 3x – 20
= 9x^2 – 3x – 20 Answer

(iv) (y^2+3/2)(y^2-3/2)

Solution: Given, (y^2+3/2)(y^2-3/2)

In the given expression
x=y^2 and y=3/2

So, polynomial can be written as follows:
(y^2+3/2)(y^2-3/2)=(y^2)^2-(3/2)^2=y^4-9/4 Answer

(v) (3 – 2x)(3 + 2x)

Answer: Given, (3 – 2x)(3 + 2x)
Here, let x = 3 and y = 2x
Using identity (x + y)(x – y) = x^2 – y^2
we get (3 – 2x)(3 + 2x) = 32 – (2x)^2
= 9 – 4x^2 Answer