Answer :
We want to factor the polynomial
[tex]$$
2x^3 + 3x^2 - 32x - 48.
$$[/tex]
Step 1. Find a Root
A good strategy is to try simple values for [tex]\( x \)[/tex] to see if they are roots. Testing [tex]\( x = 4 \)[/tex]:
[tex]$$
2(4)^3 + 3(4)^2 - 32(4) - 48 = 2(64) + 3(16) - 128 - 48 = 128 + 48 - 128 - 48 = 0.
$$[/tex]
Since the result is 0, [tex]\( x = 4 \)[/tex] is a root. This means that [tex]\( (x - 4) \)[/tex] is a factor of the polynomial.
Step 2. Divide the Polynomial by [tex]\( (x - 4) \)[/tex]
Divide
[tex]$$
2x^3 + 3x^2 - 32x - 48
$$[/tex]
by [tex]\( x - 4 \)[/tex] to obtain a quadratic. The division yields:
[tex]$$
2x^3 + 3x^2 - 32x - 48 = (x - 4)(2x^2 + 11x + 12).
$$[/tex]
Step 3. Factor the Quadratic
Now, factor the quadratic
[tex]$$
2x^2 + 11x + 12.
$$[/tex]
We look for two numbers that multiply to [tex]\( 2 \times 12 = 24 \)[/tex] and add up to [tex]\( 11 \)[/tex]. These numbers are [tex]\( 3 \)[/tex] and [tex]\( 8 \)[/tex] because [tex]\( 3 + 8 = 11 \)[/tex] and [tex]\( 3 \times 8 = 24 \)[/tex].
Rewrite the middle term:
[tex]$$
2x^2 + 11x + 12 = 2x^2 + 3x + 8x + 12.
$$[/tex]
Group the terms:
[tex]$$
(2x^2 + 3x) + (8x + 12).
$$[/tex]
Factor each group:
- From [tex]\( 2x^2 + 3x \)[/tex], factor out [tex]\( x \)[/tex]:
[tex]$$
x(2x + 3).
$$[/tex]
- From [tex]\( 8x + 12 \)[/tex], factor out [tex]\( 4 \)[/tex]:
[tex]$$
4(2x + 3).
$$[/tex]
Now, since both groups contain the common factor [tex]\( (2x + 3) \)[/tex], we have:
[tex]$$
2x^2 + 11x + 12 = (2x + 3)(x + 4).
$$[/tex]
Step 4. Write the Final Factored Form
Substitute back into the factorization:
[tex]$$
2x^3 + 3x^2 - 32x - 48 = (x - 4)(2x + 3)(x + 4).
$$[/tex]
Thus, the fully factored form of the polynomial is:
[tex]$$
\boxed{(x - 4)(x + 4)(2x + 3)}.
$$[/tex]
[tex]$$
2x^3 + 3x^2 - 32x - 48.
$$[/tex]
Step 1. Find a Root
A good strategy is to try simple values for [tex]\( x \)[/tex] to see if they are roots. Testing [tex]\( x = 4 \)[/tex]:
[tex]$$
2(4)^3 + 3(4)^2 - 32(4) - 48 = 2(64) + 3(16) - 128 - 48 = 128 + 48 - 128 - 48 = 0.
$$[/tex]
Since the result is 0, [tex]\( x = 4 \)[/tex] is a root. This means that [tex]\( (x - 4) \)[/tex] is a factor of the polynomial.
Step 2. Divide the Polynomial by [tex]\( (x - 4) \)[/tex]
Divide
[tex]$$
2x^3 + 3x^2 - 32x - 48
$$[/tex]
by [tex]\( x - 4 \)[/tex] to obtain a quadratic. The division yields:
[tex]$$
2x^3 + 3x^2 - 32x - 48 = (x - 4)(2x^2 + 11x + 12).
$$[/tex]
Step 3. Factor the Quadratic
Now, factor the quadratic
[tex]$$
2x^2 + 11x + 12.
$$[/tex]
We look for two numbers that multiply to [tex]\( 2 \times 12 = 24 \)[/tex] and add up to [tex]\( 11 \)[/tex]. These numbers are [tex]\( 3 \)[/tex] and [tex]\( 8 \)[/tex] because [tex]\( 3 + 8 = 11 \)[/tex] and [tex]\( 3 \times 8 = 24 \)[/tex].
Rewrite the middle term:
[tex]$$
2x^2 + 11x + 12 = 2x^2 + 3x + 8x + 12.
$$[/tex]
Group the terms:
[tex]$$
(2x^2 + 3x) + (8x + 12).
$$[/tex]
Factor each group:
- From [tex]\( 2x^2 + 3x \)[/tex], factor out [tex]\( x \)[/tex]:
[tex]$$
x(2x + 3).
$$[/tex]
- From [tex]\( 8x + 12 \)[/tex], factor out [tex]\( 4 \)[/tex]:
[tex]$$
4(2x + 3).
$$[/tex]
Now, since both groups contain the common factor [tex]\( (2x + 3) \)[/tex], we have:
[tex]$$
2x^2 + 11x + 12 = (2x + 3)(x + 4).
$$[/tex]
Step 4. Write the Final Factored Form
Substitute back into the factorization:
[tex]$$
2x^3 + 3x^2 - 32x - 48 = (x - 4)(2x + 3)(x + 4).
$$[/tex]
Thus, the fully factored form of the polynomial is:
[tex]$$
\boxed{(x - 4)(x + 4)(2x + 3)}.
$$[/tex]
