Question

Factor completely
4x^2-8x-21

Answer 1

To factor the quadratic expression \(4x^2 - 8x - 21\), we will look for two numbers that multiply to \(a \cdot c\) (where \(a\) is the coefficient of \(x^2\) and \(c\) is the constant term) and add to \(b\) (the coefficient of \(x\)).

1. Identify \(a\), \(b\), and \(c\):
- \(a = 4\)
- \(b = -8\)
- \(c = -21\)

2. Calculate \(a \cdot c\):
\[
a \cdot c = 4 \cdot (-21) = -84
\]

3. We need to find two numbers that multiply to \(-84\) and add to \(-8\). The numbers that work are \(-14\) and \(6\):
\[
-14 + 6 = -8
\]
\[
-14 \cdot 6 = -84
\]

4. Rewrite the middle term using these two numbers:
\[
4x^2 - 14x + 6x - 21
\]

5. Now, group the terms:
\[
(4x^2 - 14x) + (6x - 21)
\]

6. Factor out the common factors from each group:
\[
2x(2x - 7) + 3(2x - 7)
\]

7. Now factor out the common binomial factor \((2x - 7)\):
\[
(2x - 7)(2x + 3)
\]

The fully factored form of \(4x^2 - 8x - 21\) is:
\[
\boxed{(2x - 7)(2x + 3)}
\]