Question

Factor out 25x^2-20xy+3y^2

Answer 1

To factor the quadratic expression \(25x^2 - 20xy + 3y^2\), we will look for two binomials in the form \((ax + by)(cx + dy)\).

1. First, identify the coefficient values from the expression:
- \(a = 25\) (coefficient of \(x^2\))
- \(b = -20\) (coefficient of \(xy\))
- \(c = 3\) (coefficient of \(y^2\))

2. Rewrite the expression:
\(25x^2 - 20xy + 3y^2\).

3. Look for two numbers that multiply to \(a \cdot c = 25 \cdot 3 = 75\) and add up to \(b = -20\).
The numbers that satisfy this condition are \(-15\) and \(-5\):
\[
-15 \times -5 = 75 \quad \text{and} \quad -15 + -5 = -20.
\]

4. Rewrite the middle term \(-20xy\) using the numbers found:
\[
25x^2 - 15xy - 5xy + 3y^2.
\]

5. Group the terms:
\[
(25x^2 - 15xy) + (-5xy + 3y^2).
\]

6. Factor out the common factors in each group:
- From the first group \(25x^2 - 15xy\), factor out \(5x\):
\[
5x(5x - 3y).
\]
- From the second group \(-5xy + 3y^2\), factor out \(-y\):
\[
-y(5x - 3y).
\]

7. Now combine the factored terms:
\[
5x(5x - 3y) - y(5x - 3y).
\]

8. Finally, factor out the common binomial factor \((5x - 3y)\):
\[
(5x - 3y)(5x - y).
\]

Thus, the fully factored form of the expression \(25x^2 - 20xy + 3y^2\) is:
\[
(5x - 3y)(5x - y).
\]