Answer :
Sure, let's break down the factorization of both expressions step-by-step.
### Problem (a): Factorize [tex]\( x^3 + 3x^2 - 16x + 12 \)[/tex]
1. Identify Possible Rational Roots:
Using the Rational Root Theorem, consider the factors of the constant term (12) and the leading coefficient (1).
- Possible rational roots: ±1, ±2, ±3, ±4, ±6, ±12.
2. Test Possible Roots:
Substitute the possible roots into the polynomial to find an actual root. When testing [tex]\( x = 2 \)[/tex]:
[tex]\[
2^3 + 3(2)^2 - 16(2) + 12 = 8 + 12 - 32 + 12 = 0
\][/tex]
So, [tex]\( x = 2 \)[/tex] is a root.
3. Factor Out the Root:
Since [tex]\( x = 2 \)[/tex] is a root, we know [tex]\( x - 2 \)[/tex] is a factor. Use synthetic or polynomial division to divide the polynomial by [tex]\( x - 2 \)[/tex]:
[tex]\[
(x^3 + 3x^2 - 16x + 12) \div (x - 2) = x^2 + 5x - 6
\][/tex]
4. Factor the Quadratic:
Now, factor the quadratic [tex]\( x^2 + 5x - 6 \)[/tex]:
- Find two numbers that multiply to -6 and add to 5: These numbers are 6 and -1.
[tex]\[
x^2 + 5x - 6 = (x + 6)(x - 1)
\][/tex]
5. Combine Factors:
The complete factorization is:
[tex]\[
(x - 2)(x + 6)(x - 1)
\][/tex]
### Problem (b): Factorize [tex]\( 6x^3 + 19x^2 + x - 6 \)[/tex]
1. Identify Possible Rational Roots:
Consider the factors of the constant term (-6) and the leading coefficient (6).
- Possible rational roots: ±1, ±2, ±3, ±6, ±1/2, ±1/3, ±1/6.
2. Test Possible Roots:
Substitute the possible roots into the polynomial to find an actual root. After checking, [tex]\( x = -3 \)[/tex] is a root:
[tex]\[
6(-3)^3 + 19(-3)^2 + (-3) - 6 = -162 + 171 - 3 - 6 = 0
\][/tex]
3. Factor Out the Root:
Since [tex]\( x = -3 \)[/tex] is a root, we have [tex]\( x + 3 \)[/tex] as a factor. Use synthetic or polynomial division to divide the polynomial by [tex]\( x + 3 \)[/tex]:
[tex]\[
(6x^3 + 19x^2 + x - 6) \div (x + 3) = 6x^2 + x - 2
\][/tex]
4. Factor the Quadratic:
Now, factor the quadratic [tex]\( 6x^2 + x - 2 \)[/tex]:
- Find two numbers that multiply to [tex]\( 6(-2) = -12 \)[/tex] and add to 1: These are 3 and -4.
[tex]\[
6x^2 + x - 2 = (2x - 1)(3x + 2)
\][/tex]
5. Combine Factors:
The complete factorization is:
[tex]\[
(x + 3)(2x - 1)(3x + 2)
\][/tex]
And there we have the factored forms of both polynomials!
### Problem (a): Factorize [tex]\( x^3 + 3x^2 - 16x + 12 \)[/tex]
1. Identify Possible Rational Roots:
Using the Rational Root Theorem, consider the factors of the constant term (12) and the leading coefficient (1).
- Possible rational roots: ±1, ±2, ±3, ±4, ±6, ±12.
2. Test Possible Roots:
Substitute the possible roots into the polynomial to find an actual root. When testing [tex]\( x = 2 \)[/tex]:
[tex]\[
2^3 + 3(2)^2 - 16(2) + 12 = 8 + 12 - 32 + 12 = 0
\][/tex]
So, [tex]\( x = 2 \)[/tex] is a root.
3. Factor Out the Root:
Since [tex]\( x = 2 \)[/tex] is a root, we know [tex]\( x - 2 \)[/tex] is a factor. Use synthetic or polynomial division to divide the polynomial by [tex]\( x - 2 \)[/tex]:
[tex]\[
(x^3 + 3x^2 - 16x + 12) \div (x - 2) = x^2 + 5x - 6
\][/tex]
4. Factor the Quadratic:
Now, factor the quadratic [tex]\( x^2 + 5x - 6 \)[/tex]:
- Find two numbers that multiply to -6 and add to 5: These numbers are 6 and -1.
[tex]\[
x^2 + 5x - 6 = (x + 6)(x - 1)
\][/tex]
5. Combine Factors:
The complete factorization is:
[tex]\[
(x - 2)(x + 6)(x - 1)
\][/tex]
### Problem (b): Factorize [tex]\( 6x^3 + 19x^2 + x - 6 \)[/tex]
1. Identify Possible Rational Roots:
Consider the factors of the constant term (-6) and the leading coefficient (6).
- Possible rational roots: ±1, ±2, ±3, ±6, ±1/2, ±1/3, ±1/6.
2. Test Possible Roots:
Substitute the possible roots into the polynomial to find an actual root. After checking, [tex]\( x = -3 \)[/tex] is a root:
[tex]\[
6(-3)^3 + 19(-3)^2 + (-3) - 6 = -162 + 171 - 3 - 6 = 0
\][/tex]
3. Factor Out the Root:
Since [tex]\( x = -3 \)[/tex] is a root, we have [tex]\( x + 3 \)[/tex] as a factor. Use synthetic or polynomial division to divide the polynomial by [tex]\( x + 3 \)[/tex]:
[tex]\[
(6x^3 + 19x^2 + x - 6) \div (x + 3) = 6x^2 + x - 2
\][/tex]
4. Factor the Quadratic:
Now, factor the quadratic [tex]\( 6x^2 + x - 2 \)[/tex]:
- Find two numbers that multiply to [tex]\( 6(-2) = -12 \)[/tex] and add to 1: These are 3 and -4.
[tex]\[
6x^2 + x - 2 = (2x - 1)(3x + 2)
\][/tex]
5. Combine Factors:
The complete factorization is:
[tex]\[
(x + 3)(2x - 1)(3x + 2)
\][/tex]
And there we have the factored forms of both polynomials!
