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

1 answer

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: \[ (2x-3)(x+5) = 0 \]