Certainly! Let's factor the polynomial [tex]\(3x^2 - 7x - 6\)[/tex] in a step-by-step manner.
1. Identify the coefficients:
- [tex]\(a = 3\)[/tex] (coefficient of [tex]\(x^2\)[/tex])
- [tex]\(b = -7\)[/tex] (coefficient of [tex]\(x\)[/tex])
- [tex]\(c = -6\)[/tex] (constant term)
2. Calculate the discriminant [tex]\(D\)[/tex] to check if the polynomial can be factored:
[tex]\[
\text{Discriminant}, D = b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot (-6) = 49 + 72 = 121
\][/tex]
Since [tex]\(D > 0\)[/tex], the polynomial has two distinct real roots, so it can be factored over the real numbers.
3. Find the roots of the quadratic equation [tex]\(3x^2 - 7x - 6 = 0\)[/tex]:
The roots are given by the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{121}}{6}
\][/tex]
Simplifying, we get:
[tex]\[
x = \frac{7 \pm 11}{6}
\][/tex]
Therefore, the two roots are:
[tex]\[
x_1 = \frac{7 + 11}{6} = \frac{18}{6} = 3
\][/tex]
[tex]\[
x_2 = \frac{7 - 11}{6} = \frac{-4}{6} = -\frac{2}{3}
\][/tex]
4. Express the polynomial in its factored form using the roots:
If a polynomial has roots [tex]\(x = p\)[/tex] and [tex]\(x = q\)[/tex], it can be factored as [tex]\((x - p)(x - q)\)[/tex]. For [tex]\(3x^2 - 7x - 6\)[/tex], the roots are [tex]\(3\)[/tex] and [tex]\(-\frac{2}{3}\)[/tex], so the factored form is:
[tex]\[
3(x - 3)\left(x + \frac{2}{3}\right)
\][/tex]
5. Adjust the factors to include the leading coefficient (if necessary):
Since we started with a leading coefficient different from 1, we should factor out the leading coefficient [tex]\(3\)[/tex] correctly:
[tex]\[
3x^2 - 7x - 6 = 3\left(x - 3\right)\left(x + \frac{2}{3}\right)
\][/tex]
However, it's neater to express with integer coefficients:
[tex]\[
3(x - 3)(3x + 2)
\][/tex]
Thus, the polynomial [tex]\(3x^2 - 7x - 6\)[/tex] can be factored as:
[tex]\[
(3x + 2)(x - 3)
\][/tex]
