Answer :
Sure! Let's determine the value of [tex]\( m \)[/tex] for which the polynomial [tex]\( p(x) = 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] is divisible by [tex]\( 1 - 2x \)[/tex] using the Factor Theorem.
The Factor Theorem states that a polynomial [tex]\( p(x) \)[/tex] is divisible by [tex]\( x - c \)[/tex] if and only if [tex]\( p(c) = 0 \)[/tex].
In this problem, the divisor is [tex]\( 1 - 2x \)[/tex], which can be rewritten as [tex]\( -2x + 1 \)[/tex]. For simplicity, let's rewrite it in the form [tex]\( x - a \)[/tex]. Setting [tex]\( -2x + 1 = 0 \)[/tex] gives us [tex]\( x = \frac{1}{2} \)[/tex]. So, we need to find [tex]\( m \)[/tex] such that [tex]\( p\left(\frac{1}{2}\right) = 0 \)[/tex].
The polynomial is [tex]\( p(x) = 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex]. Let's now substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial and set it equal to 0:
[tex]\[ p\left( \frac{1}{2} \right) = 2\left( \frac{1}{2} \right)^4 - m\left( \frac{1}{2} \right)^3 + 4\left( \frac{1}{2} \right)^2 + 2\left( \frac{1}{2} \right) + 1 = 0 \][/tex]
Calculating each term:
[tex]\[ 2\left( \frac{1}{2} \right)^4 = 2 \left( \frac{1}{16} \right) = \frac{2}{16} = \frac{1}{8} \][/tex]
[tex]\[ -m\left( \frac{1}{2} \right)^3 = -m \left( \frac{1}{8} \right) = -\frac{m}{8} \][/tex]
[tex]\[ 4\left( \frac{1}{2} \right)^2 = 4 \left( \frac{1}{4} \right) = 1 \][/tex]
[tex]\[ 2\left( \frac{1}{2} \right) = 1 \][/tex]
[tex]\[ 1 = 1 \][/tex]
Now, substitute these values back into the polynomial equation:
[tex]\[ \frac{1}{8} - \frac{m}{8} + 1 + 1 + 1 = 0 \][/tex]
Simplify the equation:
[tex]\[ \frac{1}{8} - \frac{m}{8} + 3 = 0 \][/tex]
Multiply the entire equation by 8 to clear the fractions:
[tex]\[ 1 - m + 24 = 0 \][/tex]
Combine the constants:
[tex]\[ 25 - m = 0 \][/tex]
Solve for [tex]\( m \)[/tex]:
[tex]\[ m = 25 \][/tex]
Therefore, the value of [tex]\( m \)[/tex] for which the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] is divisible by [tex]\( 1 - 2x \)[/tex] is [tex]\( \boxed{25} \)[/tex].
The Factor Theorem states that a polynomial [tex]\( p(x) \)[/tex] is divisible by [tex]\( x - c \)[/tex] if and only if [tex]\( p(c) = 0 \)[/tex].
In this problem, the divisor is [tex]\( 1 - 2x \)[/tex], which can be rewritten as [tex]\( -2x + 1 \)[/tex]. For simplicity, let's rewrite it in the form [tex]\( x - a \)[/tex]. Setting [tex]\( -2x + 1 = 0 \)[/tex] gives us [tex]\( x = \frac{1}{2} \)[/tex]. So, we need to find [tex]\( m \)[/tex] such that [tex]\( p\left(\frac{1}{2}\right) = 0 \)[/tex].
The polynomial is [tex]\( p(x) = 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex]. Let's now substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial and set it equal to 0:
[tex]\[ p\left( \frac{1}{2} \right) = 2\left( \frac{1}{2} \right)^4 - m\left( \frac{1}{2} \right)^3 + 4\left( \frac{1}{2} \right)^2 + 2\left( \frac{1}{2} \right) + 1 = 0 \][/tex]
Calculating each term:
[tex]\[ 2\left( \frac{1}{2} \right)^4 = 2 \left( \frac{1}{16} \right) = \frac{2}{16} = \frac{1}{8} \][/tex]
[tex]\[ -m\left( \frac{1}{2} \right)^3 = -m \left( \frac{1}{8} \right) = -\frac{m}{8} \][/tex]
[tex]\[ 4\left( \frac{1}{2} \right)^2 = 4 \left( \frac{1}{4} \right) = 1 \][/tex]
[tex]\[ 2\left( \frac{1}{2} \right) = 1 \][/tex]
[tex]\[ 1 = 1 \][/tex]
Now, substitute these values back into the polynomial equation:
[tex]\[ \frac{1}{8} - \frac{m}{8} + 1 + 1 + 1 = 0 \][/tex]
Simplify the equation:
[tex]\[ \frac{1}{8} - \frac{m}{8} + 3 = 0 \][/tex]
Multiply the entire equation by 8 to clear the fractions:
[tex]\[ 1 - m + 24 = 0 \][/tex]
Combine the constants:
[tex]\[ 25 - m = 0 \][/tex]
Solve for [tex]\( m \)[/tex]:
[tex]\[ m = 25 \][/tex]
Therefore, the value of [tex]\( m \)[/tex] for which the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] is divisible by [tex]\( 1 - 2x \)[/tex] is [tex]\( \boxed{25} \)[/tex].
