5x^2y-15xy-140y

Question

Look for a GCF first, then factor the remaining trinomial

Answer 1

First, let's find the GCF of the original expression:
The coefficients of 5, 15, and 140 have a common factor of 5. The variables x^2, x, and y have a common factor of x.

GCF = 5x

Now we can factor out the GCF from the original expression:
5x(5xy - 3y - 28)

The remaining trinomial 5xy - 3y - 28 can be factored further:
5x(5xy - 3y - 28) = 5x(5xy - 20y + 17y - 28)
= 5x[5y(x - 4) + 17(y - 28)]

Therefore, the factored form of 5x^2y - 15xy - 140y is 5x(5y(x - 4) + 17(y - 28))