Answer :
To factor the polynomial
[tex]$$9x^5 + 12x^4 - 21x^3,$$[/tex]
we proceed with the following steps:
1. Factor out the Greatest Common Factor (GCF):
All three terms share a common factor. The coefficients 9, 12, and [tex]\(-21\)[/tex] have a greatest common divisor of 3, and each term also has at least [tex]\(x^3\)[/tex]. Therefore, we factor out [tex]\(3x^3\)[/tex]:
[tex]$$
9x^5 + 12x^4 - 21x^3 = 3x^3\left(\frac{9x^5}{3x^3} + \frac{12x^4}{3x^3} - \frac{21x^3}{3x^3}\right) = 3x^3\left(3x^2 + 4x - 7\right).
$$[/tex]
2. Factor the Quadratic [tex]\(3x^2 + 4x - 7\)[/tex]:
Next, we need to factor the quadratic expression [tex]\(3x^2 + 4x - 7\)[/tex]. We look for two numbers that multiply to [tex]\(3 \times (-7) = -21\)[/tex] and add up to 4. These two numbers are 7 and [tex]\(-3\)[/tex] because [tex]\(7 + (-3) = 4\)[/tex] and [tex]\(7 \times (-3) = -21\)[/tex].
We can rewrite the middle term using these two numbers:
[tex]$$
3x^2 + 4x - 7 = 3x^2 + 7x - 3x - 7.
$$[/tex]
Group the terms:
[tex]$$
(3x^2 + 7x) + (-3x - 7).
$$[/tex]
Factor by grouping:
- From the first group, factor out [tex]\(x\)[/tex]:
[tex]$$
x(3x + 7).
$$[/tex]
- From the second group, factor out [tex]\(-1\)[/tex]:
[tex]$$
-1(3x + 7).
$$[/tex]
Since both groups contain the common factor [tex]\(3x + 7\)[/tex], we factor it out:
[tex]$$
3x^2 + 7x - 3x - 7 = (3x + 7)(x - 1).
$$[/tex]
3. Write the Final Factorization:
Substitute the factored quadratic back into the expression:
[tex]$$
9x^5 + 12x^4 - 21x^3 = 3x^3 (3x^2 + 4x - 7) = 3x^3 (3x + 7)(x - 1).
$$[/tex]
Thus, the completely factored form of the given polynomial is:
[tex]$$
\boxed{3x^3 (x-1)(3x+7)}.
$$[/tex]
[tex]$$9x^5 + 12x^4 - 21x^3,$$[/tex]
we proceed with the following steps:
1. Factor out the Greatest Common Factor (GCF):
All three terms share a common factor. The coefficients 9, 12, and [tex]\(-21\)[/tex] have a greatest common divisor of 3, and each term also has at least [tex]\(x^3\)[/tex]. Therefore, we factor out [tex]\(3x^3\)[/tex]:
[tex]$$
9x^5 + 12x^4 - 21x^3 = 3x^3\left(\frac{9x^5}{3x^3} + \frac{12x^4}{3x^3} - \frac{21x^3}{3x^3}\right) = 3x^3\left(3x^2 + 4x - 7\right).
$$[/tex]
2. Factor the Quadratic [tex]\(3x^2 + 4x - 7\)[/tex]:
Next, we need to factor the quadratic expression [tex]\(3x^2 + 4x - 7\)[/tex]. We look for two numbers that multiply to [tex]\(3 \times (-7) = -21\)[/tex] and add up to 4. These two numbers are 7 and [tex]\(-3\)[/tex] because [tex]\(7 + (-3) = 4\)[/tex] and [tex]\(7 \times (-3) = -21\)[/tex].
We can rewrite the middle term using these two numbers:
[tex]$$
3x^2 + 4x - 7 = 3x^2 + 7x - 3x - 7.
$$[/tex]
Group the terms:
[tex]$$
(3x^2 + 7x) + (-3x - 7).
$$[/tex]
Factor by grouping:
- From the first group, factor out [tex]\(x\)[/tex]:
[tex]$$
x(3x + 7).
$$[/tex]
- From the second group, factor out [tex]\(-1\)[/tex]:
[tex]$$
-1(3x + 7).
$$[/tex]
Since both groups contain the common factor [tex]\(3x + 7\)[/tex], we factor it out:
[tex]$$
3x^2 + 7x - 3x - 7 = (3x + 7)(x - 1).
$$[/tex]
3. Write the Final Factorization:
Substitute the factored quadratic back into the expression:
[tex]$$
9x^5 + 12x^4 - 21x^3 = 3x^3 (3x^2 + 4x - 7) = 3x^3 (3x + 7)(x - 1).
$$[/tex]
Thus, the completely factored form of the given polynomial is:
[tex]$$
\boxed{3x^3 (x-1)(3x+7)}.
$$[/tex]
