Question 1102397
IN THIS CASE:
The terms of {{{4x^2+12x+9}}} are
{{{4x^2=(red(2x))^2}}} ,
{{{9=(green(3))^2}}} , and
{{{12x=2(red(2x))(green(3))}}} , so
{{{4x^2+12x+9=(red(2x))^2+2(red(2x))(green(3))+(green(3))^2=(red(2x)+green(3))^2}}}
 
IN GENERAL:
When you have trouble factoring a polynomial of degree 2 with a coefficient different from {{{" " +- 1}}} in front of {{{x^2}}} ,
you can split the term in {{{x}}} into two terms, and factor by parts,
like this:
{{{4x^2+12x+9=4x^2+6x+6x+9=2x(2x+3)+6x+9=2x(2x+3)+3(2x+3)=(2x+3)(2x+3)}}} .
To find how to split the term in {{{x}}},
look for factors of the product of the other two coefficients
(in this case, factors of {{{4*9=36}}} ).
Look for a pair that adds up to the coefficient of {{{x}}},
and when multiplied gives the product of the other two coefficients.
In this case {{{6*6=36}}} and {{{6+6=12}}} .