Question

Solve the quadratic by factoring.
3, x, squared, plus, 5, x, equals, minus, 3, x, minus, 4
3x
2
+5x=−3x−4

Answer 1

To solve the quadratic equation by factoring, first, we'll move all the terms to one side of the equation to set it to zero.

Starting with the original equation:

\[
3x^2 + 5x = -3x - 4
\]

Add \(3x\) and \(4\) to both sides to move them over:

\[
3x^2 + 5x + 3x + 4 = 0
\]

Combine like terms:

\[
3x^2 + 8x + 4 = 0
\]

Next, we will factor the quadratic equation. We need two numbers that multiply to \(3 \cdot 4 = 12\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(8\) (the coefficient of \(x\)).

The two numbers that fulfill this requirement are \(6\) and \(2\). We rewrite the quadratic as:

\[
3x^2 + 6x + 2x + 4 = 0
\]

Next, we can group the terms:

\[
(3x^2 + 6x) + (2x + 4) = 0
\]

Factor out the common factors in each group:

\[
3x(x + 2) + 2(x + 2) = 0
\]

Now, factor out the common binomial factor \((x + 2)\):

\[
(3x + 2)(x + 2) = 0
\]

Now, we set each factor to zero:

1. \(3x + 2 = 0\)
2. \(x + 2 = 0\)

Solving the first equation:
\[
3x + 2 = 0 \\
3x = -2 \\
x = -\frac{2}{3}
\]

Solving the second equation:
\[
x + 2 = 0 \\
x = -2
\]

Thus, the solutions to the quadratic equation are:

\[
x = -\frac{2}{3} \quad \text{and} \quad x = -2
\]