Question

Dorothy was solving the following quadratic equation by factoring: 2x^{2}+7x-15=0. Which choice shows the correctly factored form of this equation?
(2x-3) (x+5) =0
(2x-3) (x-5) =0
(5x-3) (x+2) =0
(5x-3) (x-2) =0

Answer 1

To factor the quadratic equation \(2x^2 + 7x - 15 = 0\), we start by looking for two numbers that multiply to \(2 \cdot (-15) = -30\) and add to \(7\).

The numbers that work are \(10\) and \(-3\), since:
- \(10 + (-3) = 7\)
- \(10 \cdot (-3) = -30\)

Next, we rewrite the middle term using these two numbers:
\[
2x^2 + 10x - 3x - 15 = 0
\]
Now using grouping, we can factor by grouping:
\[
(2x^2 + 10x) + (-3x - 15) = 0
\]
Factoring out the common factors in each group gives us:
\[
2x(x + 5) - 3(x + 5) = 0
\]
Now we can factor out the common binomial \((x + 5)\):
\[
(2x - 3)(x + 5) = 0
\]

So the correct factored form of the equation is \((2x - 3)(x + 5) = 0\).

Thus, the correct choice is:
\[
(2x-3)(x+5) = 0
\]