Answer :
We wish to factor the polynomial
[tex]$$
2x^3 - 3x^2 - 32x + 48.
$$[/tex]
Step 1. Check for a common factor:
We start by testing a simple candidate. Substituting [tex]$x = 4$[/tex] into the polynomial, we have
[tex]$$
2(4)^3 - 3(4)^2 - 32(4) + 48 = 128 - 48 - 128 + 48 = 0.
$$[/tex]
Since the polynomial equals zero when [tex]$x=4$[/tex], it follows that [tex]$x-4$[/tex] is a factor.
Step 2. Divide the polynomial by [tex]$x-4$[/tex]:
Divide
[tex]$$
2x^3 - 3x^2 - 32x + 48
$$[/tex]
by [tex]$x-4$[/tex]. The result of this division is a quadratic polynomial:
[tex]$$
2x^2 + 5x - 12.
$$[/tex]
Step 3. Factor the quadratic polynomial:
Now we factor
[tex]$$
2x^2 + 5x - 12.
$$[/tex]
We look for two numbers whose product is [tex]$2 \cdot (-12) = -24$[/tex] and whose sum is [tex]$5$[/tex]. The numbers [tex]$8$[/tex] and [tex]$-3$[/tex] fit the criteria because [tex]$8 \times (-3) = -24$[/tex] and [tex]$8 + (-3) = 5$[/tex]. Using these, we rewrite the middle term:
[tex]$$
2x^2 + 8x - 3x - 12.
$$[/tex]
Grouping the terms, we have
[tex]$$
(2x^2 + 8x) + (-3x - 12).
$$[/tex]
Factor out the common factors from each group:
[tex]$$
2x(x + 4) - 3(x + 4).
$$[/tex]
Noticing the common factor [tex]$(x + 4)$[/tex], we factor it out:
[tex]$$
(x+4)(2x-3).
$$[/tex]
Step 4. Write the complete factorization:
Substitute back the factor [tex]$x-4$[/tex] from Step 1. Thus, the fully factored form of the original polynomial is:
[tex]$$
2x^3 - 3x^2 - 32x + 48 = (x-4)(x+4)(2x-3).
$$[/tex]
This is the final factorization.
[tex]$$
2x^3 - 3x^2 - 32x + 48.
$$[/tex]
Step 1. Check for a common factor:
We start by testing a simple candidate. Substituting [tex]$x = 4$[/tex] into the polynomial, we have
[tex]$$
2(4)^3 - 3(4)^2 - 32(4) + 48 = 128 - 48 - 128 + 48 = 0.
$$[/tex]
Since the polynomial equals zero when [tex]$x=4$[/tex], it follows that [tex]$x-4$[/tex] is a factor.
Step 2. Divide the polynomial by [tex]$x-4$[/tex]:
Divide
[tex]$$
2x^3 - 3x^2 - 32x + 48
$$[/tex]
by [tex]$x-4$[/tex]. The result of this division is a quadratic polynomial:
[tex]$$
2x^2 + 5x - 12.
$$[/tex]
Step 3. Factor the quadratic polynomial:
Now we factor
[tex]$$
2x^2 + 5x - 12.
$$[/tex]
We look for two numbers whose product is [tex]$2 \cdot (-12) = -24$[/tex] and whose sum is [tex]$5$[/tex]. The numbers [tex]$8$[/tex] and [tex]$-3$[/tex] fit the criteria because [tex]$8 \times (-3) = -24$[/tex] and [tex]$8 + (-3) = 5$[/tex]. Using these, we rewrite the middle term:
[tex]$$
2x^2 + 8x - 3x - 12.
$$[/tex]
Grouping the terms, we have
[tex]$$
(2x^2 + 8x) + (-3x - 12).
$$[/tex]
Factor out the common factors from each group:
[tex]$$
2x(x + 4) - 3(x + 4).
$$[/tex]
Noticing the common factor [tex]$(x + 4)$[/tex], we factor it out:
[tex]$$
(x+4)(2x-3).
$$[/tex]
Step 4. Write the complete factorization:
Substitute back the factor [tex]$x-4$[/tex] from Step 1. Thus, the fully factored form of the original polynomial is:
[tex]$$
2x^3 - 3x^2 - 32x + 48 = (x-4)(x+4)(2x-3).
$$[/tex]
This is the final factorization.
