Question

How do you solve `2x^2 + 15x + 7 = 0` by factoring?

Answer 1

The solutions are `x=-1/2` and `x=-7`.

Explanation:

First, find two numbers that multiply to `2*7` (the first value times the last value), or `14`, and also add up to `15` (the middle value).

After some experimentation, you will find that these numbers are `14` and `1`. Split the `x` term into these two numbers, then factor the first and last two terms separately, then combine them. The process looks like this:

`2x^2+15x+7=0`

`2x^2+x+14x+7=0`

`color(red)x(2x+1)+14x+7=0`

`color(red)x(2x+1)+color(blue)7(2x+1)=0`

`(color(red)x+color(blue)7)(2x+1)=0`

Now, set each of the factors equal to zero and solve for `x`, and those will be the solutions:

#color(white){color(black)( (x+7=0,qquad2x+1=0), (x=-7,qquad2x=-1), (,qquadx=-1/2)):}#

You can verify these answer by seeing that these are the values where the function crosses the `x`-axis:

graph{2x^2+15x+7 [-10, 2, -25, 9]}

Answer 2

`x=-7, -1/2`

Explanation:

Solve by factoring:

`2x^2+15x+7=0`

Use the "splitting the middle term" method.

Multiply the coefficient of the first term by the constant.

`2xx7=14`

Find two numbers that when added equal `15` (the coefficient for the middle term) and when multiplied equal `14`. The numbers `1` and `14` meet the requirements.

Rewrite the equation with `14x` and `x` in place of `15x` (the middle term).

`2x^2+14x+x+7=0`

Factor out the common term in the first two terms, and the second two terms.

`2x(x+7)+(x+7)=0`

Recall that no number in front of the parentheses is understood to be `1`.

`(x+7)(2x+1)=0`

Solutions for `x`.

`x+7=0`

`x=-7`

`2x+1=0`

`2x=-1`

`x=-1/2`

`x=-7, -1/2`

graph{y=2x^2+15x+7 [-16.62, 15.42, -7.76, 8.26]}

Answer 3

• 7, and `- 1/2`

Explanation:

y = 2x^2 + 15x + 7 = 0
y = 2x^2 + 14x + x + 7 = 0
y = 2x(x +7) + (x + 7) = 0
y = (x + 7)(2x + 1) = 0
x + 7 = 0 --> x = -7
2x + 1 = 0 --> x = - 1/2