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First, we need to factor the expression [tex]\(2x^2 + 7x + 6\)[/tex].

[tex]\[
\begin{array}{l}
2x^2 + 7x + 6 \\
2x^2 + 3x + 4x + 6 \\
\left(2x^2 + 3x\right) + \left(4x + 6\right) \\
x(2x + 3) + 2(2x + 3) \\
(x + 2)(2x + 3)
\end{array}
\][/tex]

So, [tex]\(2x^2 + 7x + 6 = (x + 2)(2x + 3)\)[/tex]. Therefore, we can rewrite our original equation as:
[tex]\[
(x + 2)(2x + 3) = 0
\][/tex]

Now we can use the zero product property to solve for [tex]\(x\)[/tex].

[tex]\[
\begin{array}{l}
2x + 3 = 0 \\
2x = -3 \\
x = -\frac{3}{2}
\end{array}
\][/tex]

OR

[tex]\[
\begin{array}{c}
x + 2 = 0 \\
x = -2
\end{array}
\][/tex]



Answer :

GinnyAnswer
Certainly! Let's go through the complete solution step-by-step.

We need to factor the quadratic equation [tex]\(2x^2 + 7x + 6 = 0\)[/tex].

1. First, identify the quadratic expression: [tex]\(2x^2 + 7x + 6\)[/tex].

2. To factor it, we look for two numbers that multiply to the product of the coefficient of [tex]\(x^2\)[/tex] (which is 2) and the constant term (which is 6). These numbers must also add up to the coefficient of [tex]\(x\)[/tex] (which is 7).

- The product of 2 (coefficient of [tex]\(x^2\)[/tex]) and 6 (constant term) is [tex]\(2 \times 6 = 12\)[/tex].
- We need two numbers that multiply to 12 and add up to 7. These numbers are 3 and 4.

3. Rewrite the middle term using these two numbers:
[tex]\[ 2x^2 + 7x + 6 = 2x^2 + 3x + 4x + 6 \][/tex]

4. Group the terms:
[tex]\[ (2x^2 + 3x) + (4x + 6) \][/tex]

5. Factor by grouping:
[tex]\[ x(2x + 3) + 2(2x + 3) \][/tex]

6. Notice that [tex]\(2x + 3\)[/tex] is a common factor:
[tex]\[ (x + 2)(2x + 3) \][/tex]

So, we have factored [tex]\(2x^2 + 7x + 6\)[/tex] as [tex]\((x + 2)(2x + 3)\)[/tex].

Now, solve the factored equation [tex]\((x + 2)(2x + 3) = 0\)[/tex] using the zero product property:

7. Set each factor to zero:
[tex]\[ x + 2 = 0 \quad \text{or} \quad 2x + 3 = 0 \][/tex]

8. Solve each equation for [tex]\(x\)[/tex]:

For [tex]\(x + 2 = 0\)[/tex]:
[tex]\[ x = -2 \][/tex]

For [tex]\(2x + 3 = 0\)[/tex]:
[tex]\[ 2x = -3 \implies x = -\frac{3}{2} \implies x = -1.5 \][/tex]

Therefore, the solutions to the quadratic equation [tex]\(2x^2 + 7x + 6 = 0\)[/tex] are:
[tex]\[ x = -2 \quad \text{and} \quad x = -1.5 \][/tex]

These are the roots of the quadratic equation.

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