Question

How do you factor the expression `(2x+3)^2+9(2x+3)+14`?

Answer 1

`(2x+3)^2+9(2x+3)+14=2(2x+5)(x+5)`

Explanation:

Let `y=2x+3`

then `(2x+3)^2+9(2x+3)+14` `hArry^2+9y+14`

= `y^2+7y+2y+14`

= `y(y+7)+2(y+7)`

= `(y+2)(y+7)`

= `(2x+3+2)(2x+3+7)`

= `(2x+5)(2x+10)`

= `2(2x+5)(x+5)`

Answer 2

Exp.`=2(x+5)(2x+5).`

Explanation:

Put `(2x+3)=y` in the given exp., this will make it `y^2+9y+14.`

Now, `7xx2=14, 7+2=9.` So, we will split `9y` as `7y+2y.`

Hence, the Exp.`=y^2+7y+2y+14=y(y+7)+2(y+7) = (y+7)(y+2).`

Returning `y` as `(2x+3),` we get,

Exp. =`(2x+3+7)( 2x+3+2)=(2x+10)(2x+5)=2(x+5)(2x+5).`

Aliter:-

One can simplify the given Exp. and then factorise!