Question

Dorothy was solving the following quadratic equation by factoring: `2x^{2}+7x-15=0`.

Answer 1

To solve the quadratic equation \(2x^2 + 7x - 15 = 0\) by factoring, we can first look for two numbers that multiply to \(2 \cdot (-15) = -30\) (the product of the coefficient of \(x^2\) and the constant term) and add up to \(7\) (the coefficient of \(x\)).

The numbers that satisfy these conditions are \(10\) and \(-3\) since:
- \(10 \cdot (-3) = -30\)
- \(10 + (-3) = 7\)

Now, we can rewrite the middle term \(7x\) using these two numbers:

\[
2x^2 + 10x - 3x - 15 = 0
\]

Next, we can factor by grouping:

1. Group the first two terms and the last two terms:
\[
(2x^2 + 10x) + (-3x - 15) = 0
\]

2. Factor out the common factors in each group:
\[
2x(x + 5) - 3(x + 5) = 0
\]

3. Now, factor out the common binomial factor \((x + 5)\):
\[
(2x - 3)(x + 5) = 0
\]

Now, we can set each factor equal to zero to find the solutions:

1. For \(2x - 3 = 0\):
\[
2x = 3 \implies x = \frac{3}{2}
\]

2. For \(x + 5 = 0\):
\[
x = -5
\]

Thus, the solutions to the quadratic equation \(2x^2 + 7x - 15 = 0\) are:
\[
x = \frac{3}{2} \quad \text{and} \quad x = -5
\]