Question

3 fill in the blank 4 points
consider the function $f(x)=x^{4}-13x^{3}+42x^{2}-8x+k$.
what is the value of k that makes (x-2) a factor of f(x)?
k = type your answer...
if (x-2) is a factor of f(x) and 4 is a zero of a function, what are the three other factors of f(x)?
type your answer... type your answer... type your answer...

Explanation:

Step1: Apply Factor Theorem

If $(x-2)$ is a factor, $f(2)=0$.
$f(2) = 2^4 -13(2)^3 +42(2)^2 -8(2)+k=0$

Step2: Calculate $f(2)$

$16 - 104 + 168 - 16 + k = 0$
$64 + k = 0$
$k = -80$

Step3: Define full polynomial

$f(x) = x^4 -13x^3 +42x^2 -8x -80$

Step4: Factor out $(x-2)$

Use polynomial division or synthetic division:
$\frac{x^4 -13x^3 +42x^2 -8x -80}{x-2}=x^3 -11x^2 +20x +40$

Step5: Use root $x=4$ on cubic

Since 4 is a zero, factor out $(x-4)$ from $x^3 -11x^2 +20x +40$:
$\frac{x^3 -11x^2 +20x +40}{x-4}=x^2 -7x -20$

Step6: Factor quadratic

$x^2 -7x -20=(x+4)(x-5)$

Step7: List all factors

The three other factors are $(x-4)$, $(x+4)$, $(x-5)$

Answer:

$k = -80$
$(x-4)$, $(x+4)$, $(x-5)$