Factor completely
2x^3y-12x^2y-14xy
2x^2y-6xy-7xy
could check this problem
2x^3y-12x^2y-14xy
= 2xy(x^2 - 6x - 7)
Was your second line supposed to be the answer?
If so, Keaton, you have a bit of work ahead of you.
Certainly! Let's factor the expression: 2x^3y - 12x^2y - 14xy
First, let's look for common factors among the terms: 2x^3y, -12x^2y, -14xy.
The common factor among the three terms is "2xy". We can factor it out:
2xy(x^2 - 6x - 7)
Now we need to factor the remaining quadratic expression, x^2 - 6x - 7:
To factorize the quadratic expression further, we need to find two numbers that multiply to give -7 and add up to -6 (the coefficient of x).
The numbers that satisfy these conditions are -7 and 1. So we can rewrite -6x as -7x + x:
2xy(x^2 - 7x + x - 7)
Now, we can group the terms:
2xy((x^2 - 7x) + (x - 7))
Next, we can factor out the common factor in each pair of terms:
2xy(x(x - 7) + 1(x - 7))
Finally, we have:
2xy(x + 1)(x - 7)
So, the completely factored expression is 2xy(x + 1)(x - 7).