Question 280358
Let x = the first even integer. Then x+2 = the second even integer. Their product would be x(x+2). This is 120:
{{{x(x+2) = 120}}}
Now we solve for x. Start by simplifying the left side:
{{{x^2+2x = 120}}}
Since this is a quadratic equation we want one side to be zero. So subtract 120 from each side:
{{{x^2+2x - 120 = 0}}}
Now we factor (or use the Quadratic Formula):
{{{(x+12)(x-10) = 0}}}
By the Zero Product Property we know that one of these factors must be zero:
x+12 = 0  or x-10 = 0
Solving these we get:
x = -12 or x = 10
That make the second even integer...
x+2 = -12 + 2 = -10 or x+2 = 10 + 2 = 12<br>
So there are two pairs of consecutive even integers whose product is 120:
-12 and -10 or 10 and 12