Question

Factor completely
6x^2-7x-10

Answer 1

To factor the quadratic expression \(6x^2 - 7x - 10\), we can use the method of factoring by grouping. First, we look for two numbers that multiply to \(ac\) (where \(a\) is the coefficient of \(x^2\) and \(c\) is the constant term) and add to \(b\) (the coefficient of \(x\)).

For the quadratic \(6x^2 - 7x - 10\):
- \(a = 6\)
- \(b = -7\)
- \(c = -10\)

Next, we calculate \(ac = 6 \times -10 = -60\).

Now we need to find two numbers that multiply to \(-60\) and add up to \(-7\). The numbers \(-12\) and \(5\) satisfy these conditions:
- \(-12 \cdot 5 = -60\)
- \(-12 + 5 = -7\)

Now we can rewrite the middle term \(-7x\) using \(-12x\) and \(5x\):

\[
6x^2 - 12x + 5x - 10
\]

Next, we group the terms:

\[
(6x^2 - 12x) + (5x - 10)
\]

We factor out the greatest common factor from each group:

\[
6x(x - 2) + 5(x - 2)
\]

Now we can factor out the common binomial factor \((x - 2)\):

\[
(6x + 5)(x - 2)
\]

Thus, the completely factored form of \(6x^2 - 7x - 10\) is:

\[
\boxed{(6x + 5)(x - 2)}
\]