The polynomial function f is defined as f(c)=(c-k)(c^(2)-4c+4) where k is a constant. The value 2 is a zero of f. What is the remainder of f(c) when divided by (c-2)?

Given Information: We are given that 2 is a zero of the polynomial function f(c). This means that f(2) should equal 0. Let's substitute c = 2 into the function to find the value of k. f(2) = (2 - k)(2^2 - 4*2 + 4)Substitute and Calculate: Now we simplify the expression by calculating the square and the other operations inside the parentheses. f(2) = (2 - k)(4 - 8 + 4)Simplify Expression: Further simplifying the expression inside the parentheses gives us: f(2) = (2 - k)(0) Since anything multiplied by 0 is 0, we have: f(2) = 0Further Simplification: Since f(2) = 0 for the zero of the polynomial, we don't need to solve for k because the value of k will not affect the remainder when f(c) is divided by (c-2). The remainder theorem tells us that the remainder of a polynomial f(c) when divided by (c - a) is simply f(a). Therefore, the remainder of f(c) when divided by (c-2) is f(2).Final Result: We have already established that f(2) = 0, so the remainder of f(c) when divided by (c-2) is 0.

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The polynomial function
f is defined as

f(c)=(c-k)(c^(2)-4c+4)
where
k is a constant. The value 2 is a zero of
f. What is the remainder of
f(c) when divided by
(c-2)?

The polynomial function $f$ is defined as $\newline$ $f(c)=(c-k)\left(c^{2}-4 c+4\right)$ $\newline$ where $k$ is a constant. The value $2$ is a zero of $f$ . What is the remainder of $f(c)$ when divided by $(c-2)$ ?

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Algebra 2

Composition of linear and quadratic functions: find a value

Full solution

Q. The polynomial function

f

is defined as

\newline

f(c)=(c-k)\left(c^{2}-4 c+4\right)

\newline

where

k

is a constant. The value

2

is a zero of

f

. What is the remainder of

f(c)

when divided by

(c-2)

Given Information: We are given that $2$ is a zero of the polynomial function $f(c)$ . This means that $f(2)$ should equal $0$ . Let's substitute $c = 2$ into the function. $\newline$ $f(2) = (2 - k)(2^2 - 4\times2 + 4)$
Substitute and Calculate: Now we simplify the expression by calculating the square and the other operations inside the parentheses. $\newline$ $f(2) = (2 - k)(4 - 8 + 4)$
Simplify Expression: Further simplifying the expression inside the parentheses gives us: $\newline$ $f(2) = (2 - k)(0)$ $\newline$ Since anything multiplied by $0$ is $0$ , we have: $\newline$ $f(2) = 0$
Further Simplification: Since $f(2) = 0$ for the zero of the polynomial, we don't need to solve for $k$ because the value of $k$ will not affect the remainder when $f(c)$ is divided by $(c-2)$ . The remainder theorem tells us that the remainder of a polynomial $f(c)$ when divided by $(c - a)$ is simply $f(a)$ . Therefore, the remainder of $f(c)$ when divided by $(c-2)$ is $0$ .
Final Result: We have already established that $f(2) = 0$ , so the remainder of $f(c)$ when divided by $(c-2)$ is $0$ .