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

2 Answers
Jun 26, 2016

#(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)#

Jun 26, 2016

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!