Answer :
Let's factor the polynomial [tex]\(9x^3 + 36x^2 - x - 4\)[/tex] completely. We seek to express it as a product of simpler polynomials.
Given: [tex]\(9x^3 + 36x^2 - x - 4\)[/tex]
We'll proceed step-by-step to factorize it.
1. Identify Possible Rational Roots:
For a polynomial [tex]\(ax^n + bx^{n-1} + \ldots + k\)[/tex], possible rational roots can be factors of the constant term divided by factors of the leading coefficient. In this case, constant term = [tex]\(-4\)[/tex] and leading coefficient = [tex]\(9\)[/tex].
2. Trial and Error for Rational Roots:
We test various potential rational roots:
- Factors of [tex]\(-4\)[/tex] are [tex]\(\pm 1, \pm 2, \pm 4\)[/tex].
- Factors of [tex]\(9\)[/tex] are [tex]\(\pm 1, \pm 3, \pm 9\)[/tex].
We test each combination, for example, [tex]\(\pm \frac{1}{1}, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm \frac{2}{1}, \pm \frac{4}{1}, etc.\)[/tex].
3. Synthetic Division:
We find through synthetic division or evaluating the polynomial that [tex]\(x = -4\)[/tex] is a root.
Dividing [tex]\(9x^3 + 36x^2 - x - 4\)[/tex] by [tex]\((x + 4)\)[/tex], we obtain the quotient [tex]\((3x^2 + 3x - 1)\)[/tex].
4. Factor the Quadratic Polynomial [tex]\(3x^2 + 3x - 1\)[/tex]:
Next, we factorize [tex]\(3x^2 + 3x - 1\)[/tex]. Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], we find:
[tex]\[ x = \frac{-3 \pm \sqrt{(3)^2 - 4(3)(-1)}}{2(3)} = \frac{-3 \pm \sqrt{9 + 12}}{6} = \frac{-3 \pm \sqrt{21}}{6} = \frac{-3 \pm 3\sqrt{21}}{6} = \frac{-3 \pm \sqrt{21}}{6} = \frac{-1 \pm \sqrt{21}}{6} \][/tex]
Therefore, [tex]\(3x^2 + 3x - 1\)[/tex] factors into [tex]\((3x - 1)(x + \frac{1}{3}) \)[/tex].
However, since we are dealing with integers in the factors, rewritting [tex]\(x + \frac{1}{3}\)[/tex] in the integer factor form, we get [tex]\((3x-1)\)[/tex] and [tex]\((x+4)\)[/tex].
Putting it all together, the polynomial [tex]\(9x^3 + 36x^2 - x - 4\)[/tex] factors completely as:
[tex]\[ (3x - 1)(3x + 1)(x + 4) \][/tex]
So, the correct answer is:
[tex]\[ (3x + 1)(3x - 1)(x + 4) \][/tex]
Given: [tex]\(9x^3 + 36x^2 - x - 4\)[/tex]
We'll proceed step-by-step to factorize it.
1. Identify Possible Rational Roots:
For a polynomial [tex]\(ax^n + bx^{n-1} + \ldots + k\)[/tex], possible rational roots can be factors of the constant term divided by factors of the leading coefficient. In this case, constant term = [tex]\(-4\)[/tex] and leading coefficient = [tex]\(9\)[/tex].
2. Trial and Error for Rational Roots:
We test various potential rational roots:
- Factors of [tex]\(-4\)[/tex] are [tex]\(\pm 1, \pm 2, \pm 4\)[/tex].
- Factors of [tex]\(9\)[/tex] are [tex]\(\pm 1, \pm 3, \pm 9\)[/tex].
We test each combination, for example, [tex]\(\pm \frac{1}{1}, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm \frac{2}{1}, \pm \frac{4}{1}, etc.\)[/tex].
3. Synthetic Division:
We find through synthetic division or evaluating the polynomial that [tex]\(x = -4\)[/tex] is a root.
Dividing [tex]\(9x^3 + 36x^2 - x - 4\)[/tex] by [tex]\((x + 4)\)[/tex], we obtain the quotient [tex]\((3x^2 + 3x - 1)\)[/tex].
4. Factor the Quadratic Polynomial [tex]\(3x^2 + 3x - 1\)[/tex]:
Next, we factorize [tex]\(3x^2 + 3x - 1\)[/tex]. Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], we find:
[tex]\[ x = \frac{-3 \pm \sqrt{(3)^2 - 4(3)(-1)}}{2(3)} = \frac{-3 \pm \sqrt{9 + 12}}{6} = \frac{-3 \pm \sqrt{21}}{6} = \frac{-3 \pm 3\sqrt{21}}{6} = \frac{-3 \pm \sqrt{21}}{6} = \frac{-1 \pm \sqrt{21}}{6} \][/tex]
Therefore, [tex]\(3x^2 + 3x - 1\)[/tex] factors into [tex]\((3x - 1)(x + \frac{1}{3}) \)[/tex].
However, since we are dealing with integers in the factors, rewritting [tex]\(x + \frac{1}{3}\)[/tex] in the integer factor form, we get [tex]\((3x-1)\)[/tex] and [tex]\((x+4)\)[/tex].
Putting it all together, the polynomial [tex]\(9x^3 + 36x^2 - x - 4\)[/tex] factors completely as:
[tex]\[ (3x - 1)(3x + 1)(x + 4) \][/tex]
So, the correct answer is:
[tex]\[ (3x + 1)(3x - 1)(x + 4) \][/tex]