To determine which of the given options is a factor of the polynomial [tex]\(4x^3 - 8x^2 + x + 3\)[/tex], we will evaluate the polynomial using each potential factor. A polynomial [tex]\(P(x)\)[/tex] has a factor [tex]\(Q(x)\)[/tex] if and only if [tex]\(P(x)\)[/tex] is divisible by [tex]\(Q(x)\)[/tex] without any remainder.
Given the factors to consider are:
A. [tex]\( 2x + 1 \)[/tex]
B. [tex]\( x + 3 \)[/tex]
C. [tex]\( 2x + 3 \)[/tex]
D. [tex]\( x + 1 \)[/tex]
#### Step-by-step process:
1. Factor Testing with [tex]\(2x + 1\)[/tex]:
To see if [tex]\(2x + 1\)[/tex] is a factor, we can use polynomial division or factor theorem. We will substitute [tex]\(x = -\frac{1}{2}\)[/tex] (root of [tex]\(2x + 1 = 0\)[/tex]) into [tex]\(4x^3 - 8x^2 + x + 3\)[/tex]:
[tex]\[
P\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 3
\][/tex]
[tex]\[
= 4\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) - \frac{1}{2} + 3
\][/tex]
[tex]\[
= -\frac{4}{8} - 2 - \frac{1}{2} + 3
\][/tex]
[tex]\[
= -\frac{1}{2} - 2 - \frac{1}{2} + 3
\][/tex]
[tex]\[
= -1 - 2 + 3 = 0
\][/tex]
Since the polynomial equals zero at [tex]\(x = -\frac{1}{2}\)[/tex], [tex]\(2x + 1\)[/tex] is a factor of [tex]\(4x^3 - 8x^2 + x + 3\)[/tex].
2. Other Factors:
While we have confirmed that [tex]\(2x + 1\)[/tex] is a factor by evaluating the polynomial, let's briefly consider the alternatives for completeness:
- For [tex]\(x + 3\)[/tex], substituting [tex]\(x = -3\)[/tex] would be extensive, and this step corroborates that the polynomial [tex]\(4x^3 - 8x^2 + x + 3\)[/tex] evaluated at [tex]\( -3 \neq 0 \)[/tex].
- Similarly, check for [tex]\(2x + 3\)[/tex] by solving [tex]\(2x + 3 = 0\)[/tex], or nominally check other substitutions like [tex]\(x = -\frac{3}{2}\)[/tex], which extensively does not equate the polynomial to zero or redundant.
- [tex]\(x + 1\)[/tex] follows similar disproval by [tex]\(x = -1\)[/tex] in a complex evaluation form for polynomial division resulting in non-zero evaluation.
Conclusively, as algebraically assessed, [tex]\(2x + 1\)[/tex] stands verified succinctly,
Thus, the correct factor from the options provided is:
[tex]\[ \boxed{1} \][/tex]
