Certainly! Let’s factorize the quadratic polynomial [tex]\(3x^2 + 22x - 16\)[/tex].
To factorize [tex]\(3x^2 + 22x - 16\)[/tex], we want to express it in the form [tex]\((ax + b)(cx + d)\)[/tex].
1. The polynomial given is [tex]\(3x^2 + 22x - 16\)[/tex].
2. First, look at the constant term [tex]\(-16\)[/tex], and the coefficient of the [tex]\(x^2\)[/tex] term, which is 3.
3. We need to find two numbers that multiply to [tex]\((3 \times -16) = -48\)[/tex] and add to [tex]\(22\)[/tex] (coefficient of [tex]\(x\)[/tex]).
4. Those two numbers are 24 and -2, because [tex]\(24 \cdot (-2) = -48\)[/tex] and [tex]\(24 + (-2) = 22\)[/tex].
5. Rewrite the middle term, [tex]\(22x\)[/tex], using 24 and -2:
[tex]\[
3x^2 + 24x - 2x - 16
\][/tex]
6. Group the terms to factor by grouping:
[tex]\[
(3x^2 + 24x) + (-2x - 16)
\][/tex]
7. Factor [tex]\(3x\)[/tex] out of the first group, and [tex]\(-2\)[/tex] out of the second group:
[tex]\[
3x(x + 8) - 2(x + 8)
\][/tex]
8. Notice that [tex]\((x + 8)\)[/tex] is a common factor:
[tex]\[
(x + 8)(3x - 2)
\][/tex]
Therefore, the complete factorization of [tex]\(3 x^2 + 22 x - 16\)[/tex] is [tex]\((x + 8)(3x - 2)\)[/tex].
