To solve the equation [tex]\(2x^2 - 6x = -4\)[/tex] by factoring, follow these steps:
1. Rewrite the equation in standard quadratic form:
The standard quadratic form is [tex]\(ax^2 + bx + c = 0\)[/tex]. Start by moving all terms to one side of the equation:
[tex]\[
2x^2 - 6x + 4 = 0
\][/tex]
2. Factor the quadratic equation:
First, we look for two numbers that multiply to [tex]\((2 \cdot 4 = 8)\)[/tex] and add up to [tex]\(-6\)[/tex]. The correct pair of numbers is [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex]. So we can rewrite the middle term [tex]\(-6x\)[/tex] as [tex]\(-4x - 2x\)[/tex]:
[tex]\[
2x^2 - 4x - 2x + 4 = 0
\][/tex]
Next, group the terms to factor by grouping:
[tex]\[
(2x^2 - 4x) - (2x - 4) = 0
\][/tex]
Factor out the greatest common factor from each group:
[tex]\[
2x(x - 2) - 2(x - 2) = 0
\][/tex]
Notice that [tex]\((x - 2)\)[/tex] is common in both groups, so we factor it out:
[tex]\[
(2x - 2)(x - 2) = 0
\][/tex]
3. Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
We now have two factors which can be set to zero:
[tex]\[
2x - 2 = 0 \quad \text{or} \quad x - 2 = 0
\][/tex]
Solve each equation for [tex]\(x\)[/tex]:
[tex]\[
\begin{aligned}
2x - 2 &= 0 \\
2x &= 2 \\
x &= 1
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
x - 2 &= 0 \\
x &= 2
\end{aligned}
\][/tex]
4. Final solutions:
The solutions to the equation [tex]\(2x^2 - 6x + 4 = 0\)[/tex] are:
[tex]\[
x = 1, \quad x = 2
\][/tex]
Therefore, the solutions are:
[tex]\[ x = \boxed{1} \][/tex]
[tex]\[ x = \boxed{2} \][/tex]
