To determine the value of [tex]\( p \)[/tex] so that the polynomial [tex]\( 4x^4 - 12x^3 + 13x^2 - 8x + p \)[/tex] is divisible by [tex]\( 2x - 1 \)[/tex], we need to use the property of polynomial division which states that for a polynomial [tex]\( f(x) \)[/tex] to be divisible by [tex]\( x - a \)[/tex], [tex]\( f(a) \)[/tex] must be zero.
Here, our divisor is [tex]\( 2x - 1 \)[/tex]. Let’s determine the value of [tex]\( x \)[/tex] (denoted as [tex]\( a \)[/tex]) that makes [tex]\( 2x - 1 = 0 \)[/tex]:
[tex]\[ 2x - 1 = 0 \][/tex]
[tex]\[ 2x = 1 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
We substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial [tex]\( 4x^4 - 12x^3 + 13x^2 - 8x + p \)[/tex] and set the result equal to zero.
First, plug [tex]\( \frac{1}{2} \)[/tex] into each term and calculate:
[tex]\[ 4 \left(\frac{1}{2}\right)^4 = 4 \left(\frac{1}{16}\right) = \frac{4}{16} = \frac{1}{4} \][/tex]
[tex]\[ -12 \left(\frac{1}{2}\right)^3 = -12 \left(\frac{1}{8}\right) = -\frac{12}{8} = -\frac{3}{2} \][/tex]
[tex]\[ 13 \left(\frac{1}{2}\right)^2 = 13 \left(\frac{1}{4}\right) = \frac{13}{4} \][/tex]
[tex]\[ -8 \left(\frac{1}{2}\right) = -8 \left(\frac{1}{2}\right) = -4 \][/tex]
Now, sum these values and equate it to zero to solve for [tex]\( p \)[/tex]:
[tex]\[
\frac{1}{4} - \frac{3}{2} + \frac{13}{4} - 4 + p = 0
\][/tex]
Convert all terms to have a common denominator, which is 4:
[tex]\[
\frac{1}{4} - \frac{3 \cdot 2}{4} + \frac{13}{4} - \frac{4 \cdot 4}{4} + p = 0
\][/tex]
[tex]\[
\frac{1}{4} - \frac{6}{4} + \frac{13}{4} - \frac{16}{4} + p = 0
\][/tex]
Combine the fractions:
[tex]\[
\frac{1 - 6 + 13 - 16}{4} + p = 0
\][/tex]
[tex]\[
\frac{1 - 6 + 13 - 16}{4} + p = 0
\][/tex]
[tex]\[
\frac{-8}{4} + p = 0
\][/tex]
[tex]\[
-2 + p = 0
\][/tex]
Thus,
[tex]\[
p = 2
\][/tex]
Therefore, the value of [tex]\( p \)[/tex] is [tex]\( \boxed{2} \)[/tex].
