Question 862077
{{{(x^2+7x+12)/(2x^2-6x-36)}}}
Factor out a 2 on the bottom
{{{(x^2+7x+12)/(2(x^2-3x-18))}}}
To factor (x^2-3x-18), find the numbers that multiply to -18 and add to -3. These numbers are -6 and 3.
{{{(x^2+7x+12)/(2(x-6)(x+3))}}}
To factor (x^2+7x+12), find the numbers that multiply to 12 and add to 7. These numbers are 3 and 4.
{{{(x+3)(x+4)/2(x-6)(x+3)}}}
Because x+3 is on both sides of the fraction, you can cancel those out.
{{{(x+4)/2(x-6)}}}
This equation is now fully factored.