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

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Answer 1 · 2016-06-26T05:22:08

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

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 · 2016-06-26T05:32:08

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

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!

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