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



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