Question 344967
{{{(x+5)(x+3)+5x+25=0}}}
The long way:
Simplify the left side:
{{{x^2 +8x +15 + 5x + 25 = 0}}}
{{{x^2 + 13x + 40 = 0}}}
Factor:
{{{(x+8)(x+5) = 0}}}
Solve:
From the Zero Product property we know that one of the factors must be zero:
{{{x+8 = 0}}} or {{{x+5 = 0}}}
Solving these we get:
{{{x = -8}}} or {{{x = -5}}}<br>
{{{(x+5)(x+3)+5x+25=0}}}
The short way is based on noticing that (x+5) is already a factor of (x+5)(x+3) and that it is also a factor of 5x+25. This means we can factor by grouping:
{{{(x+5)(x+3)+5(x+5) = 0}}}
{{{(x+5)((x+3)+5) = 0}}}
{{{(x+5)(x+8) = 0}}}
And the rest is the same as the above.