Question

Pls help :)) this question is confusing for me :)

Answer 1

Given:

$y^{2}(2 y-5)-8 y+20$

To find:

The product of the expression.

Solution:

$y^{2}(2 y-5)-8 y+20=y^{2}(2 y-5)+(-8 y+20)$

Take out -4 as a common factor in Last two terms.

                                $=y^{2}(2 y-5)-4(2 y-5)$

Make sure the terms in both brackets must be same.

Take out common factor (2y - 5) from both terms.

                                $=(2 y-5)(y^{2}-4)$

4 can be written as 2².

                                $=(2 y-5)\left(y^{2}-2^2\right)$

Using the identity: $(a^2-b^2)=(a-b)(a+b)$

                                $=(2 y-5)(y-2)(y+2)$.

The product is $(2 y-5)(y-2)(y+2)$.