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Question
let $f(x) = 2x^2 - x + 1$ and let $g(x) = x - 4$\which statements are true?\select all that apply.\\a. if $f(x)$ is divided by $g(x)$, the remainder is 29.\\b. one root of $f(x)$ is 20.\\c. one root of $f(x)$ is -4.\\d. $g(x)$ is a factor of $f(x)$\\e. $g(x)$ is not a factor of $f(x)$
To determine the validity of each statement, we use the Remainder Theorem and analyze the roots of \( f(x) \). The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x - a \), the remainder is \( f(a) \). For \( g(x) = x - 4 \), we evaluate \( f(4) \) to find the remainder when \( f(x) \) is divided by \( g(x) \). We also check if \( g(x) \) is a factor by seeing if \( f(4) = 0 \), and analyze the roots of \( f(x) \) by solving \( 2x^2 - x + 1 = 0 \).
Step 1: Evaluate \( f(4) \) for Statement A
Using the Remainder Theorem, substitute \( x = 4 \) into \( f(x) = 2x^2 - x + 1 \):
\[
f(4) = 2(4)^2 - 4 + 1 = 2(16) - 4 + 1 = 32 - 4 + 1 = 29
\]
So, the remainder when \( f(x) \) is divided by \( g(x) \) is 29. Thus, Statement A is true.
Step 2: Check if \( g(x) \) is a factor (Statement D and E)
For \( g(x) = x - 4 \) to be a factor of \( f(x) \), \( f(4) \) must be 0. But we found \( f(4) = 29
eq 0 \), so \( g(x) \) is not a factor of \( f(x) \). Thus, Statement D is false, and Statement E is true.
Step 3: Analyze roots of \( f(x) \) (Statement B and C)
Solve the quadratic equation \( 2x^2 - x + 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = 1 \):
\[
\text{Discriminant} = (-1)^2 - 4(2)(1) = 1 - 8 = -7
\]
Since the discriminant is negative, the roots are complex (not real). Thus, neither 20 nor -4 can be roots (as they are real numbers), so Statements B and C are false.
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A. if \( f(x) \) is divided by \( g(x) \), the remainder is 29
E. \( g(x) \) is not a factor of \( f(x) \)