To factor the polynomial [tex]\(x^2 + 7x + 12\)[/tex], we need to find two binomials whose product equals the given polynomial.
We start with the form:
[tex]\[
(x + a)(x + b)
\][/tex]
Expanding this product, we get:
[tex]\[
(x + a)(x + b) = x^2 + (a + b)x + ab
\][/tex]
By comparing this with the original polynomial [tex]\(x^2 + 7x + 12\)[/tex], we can see that:
[tex]\[
a + b = 7 \quad \text{and} \quad ab = 12
\][/tex]
Our task is to find two numbers, [tex]\(a\)[/tex] and [tex]\(b\)[/tex], that satisfy these conditions. Let's list pairs of numbers that multiply to 12 and check their sum:
[tex]\[
\begin{array}{l}
1 \cdot 12 = 12 \quad \text{and} \quad 1 + 12 = 13 \\
2 \cdot 6 = 12 \quad \text{and} \quad 2 + 6 = 8 \\
3 \cdot 4 = 12 \quad \text{and} \quad 3 + 4 = 7 \\
\end{array}
\][/tex]
Among these pairs, the pair that both multiplies to 12 and adds to 7 is [tex]\(3\)[/tex] and [tex]\(4\)[/tex]. Therefore, we have identified [tex]\(a = 3\)[/tex] and [tex]\(b = 4\)[/tex].
Thus, the factored form of the polynomial [tex]\(x^2 + 7x + 12\)[/tex] is:
[tex]\[
(x + 3)(x + 4)
\][/tex]
We can always verify our solution by expanding the factors back out:
[tex]\[
(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12
\][/tex]
Since this matches the original polynomial, our factorization is confirmed correct. Therefore, the factored form of [tex]\(x^2 + 7x + 12\)[/tex] is:
[tex]\[
(x + 3)(x + 4)
\][/tex]
