Question

factor completely
2x^2-9x+7

Answer 1

To factor the quadratic expression \(2x^2 - 9x + 7\) completely, we can use the method of factoring by grouping or trial and error to identify 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 the coefficients:
- \(a = 2\)
- \(b = -9\)
- \(c = 7\)

2. Compute \(a \cdot c\):
\[
a \cdot c = 2 \cdot 7 = 14
\]

3. We are looking for two numbers that multiply to \(14\) and add to \(-9\).

The numbers that satisfy these conditions are \(-7\) and \(-2\) since:
\(-7 \times -2 = 14\)
\(-7 + -2 = -9\)

4. Rewrite the middle term \(-9x\) using \(-7x\) and \(-2x\):
\[
2x^2 - 7x - 2x + 7
\]

5. Now, group the terms:
\[
(2x^2 - 7x) + (-2x + 7)
\]

6. Factor out the common factors in each group:
\[
x(2x - 7) - 1(2x - 7)
\]

7. Now factor out the common binomial factor \((2x - 7)\):
\[
(2x - 7)(x - 1)
\]

So, the completely factored form of \(2x^2 - 9x + 7\) is:
\[
\boxed{(2x - 7)(x - 1)}
\]