Factor the expression completely.

Question

x2y3 - x5y4

Answer 1

Try expanding the equation to help you visualize what is happening:

1. For example, x² = x * x and y³ = y * y * y
2. Therefore x²y³ = x * x * y * y * y
3. Applying this to the entire expression, we get: x * x * y * y * y - x * x * x * x * x * y * y * y * y

When factoring, we look for anything our terms have in common:

1. For example, both terms have 2 x's.
2. Pulling out those x's, we get x * x * (x * x * y * y * y - x * x * x * x * x * y * y * y * y)
3. Or more cleanly, x * x * (y * y * y - x * x * x * y * y * y * y)
4. Now, only one term has any x's. (The second term still has 3)
5. Both terms still have y's. (The first term has 3 y's and the second term has 4)
6. Since they both have at least 3 y's, we can pull those out as well
7. x * x * y * y * y * (y * y * y - x * x * x * y * y * y * y)
8. Or more cleanly, x * x * y * y * y * (1 - x * x * x * y)

Returning to the use of exponents, we have x²y³(1 - x³y)

Answer 2

I assume this is x²y³ - x⁵y⁴ so lets pull out like terms:

x²y³(1 - x³y) and this does not factor any further.