Question $x-2 ; f(x)=x^{4}-3 x^{3}+2 x^{2}+8 x-16$

Studdy Solution

STEP 1

What is this asking? We're trying to figure out if $x-2$ is a factor of the polynomial $f(x) = x^4 - 3x^3 + 2x^2 + 8x - 16$ .
If it is, we'll find out what we get when we divide $f(x)$ by $x-2$ . Watch out! Remember, if $x-2$ is a factor, the remainder when we divide should be zero!
Also, be super careful with those signs when we're doing polynomial long division or synthetic division.

STEP 2

1. Check if $x-2$ is a factor
2. Perform polynomial long division

STEP 3

Let's set our potential factor, $x-2$ , equal to zero and solve for $x$ : $x - 2 = 0$ $x = 2$

STEP 4

Now, we'll plug in this value of $x=2$ into our polynomial $f(x)$ : $f(2) = (2)^4 - 3(2)^3 + 2(2)^2 + 8(2) - 16$ $f(2) = 16 - 3(8) + 2(4) + 16 - 16$ $f(2) = 16 - 24 + 8 + 16 - 16$ $f(2) = 0$

STEP 5

Since $f(2) = 0$ , the Factor Theorem tells us that yes, $x-2$ is a factor of $f(x)$ !
Woohoo!

STEP 6

Alright, let's divide $f(x) = x^4 - 3x^3 + 2x^2 + 8x - 16$ by $x-2$ .

STEP 7

$\begin{array}{cccc} x^3 & -x^2 & +0x & +8 \\ x-2 & x^4 & -3x^3 & +2x^2 & +8x & -16 \\ x^4 & -2x^3 \\ 0 & -x^3 & +2x^2 \\ -x^3 & +2x^2 \\ 0 & 0 & +8x & -16\\ +8x & -16\\ 0 & 0\\ \end{array}$

STEP 8

The quotient is $x^3 - x^2 + 8$ , and the remainder is zero, as expected!

STEP 9

So, $x-2$ is indeed a factor of $x^4 - 3x^3 + 2x^2 + 8x - 16$ , and when we divide, we get $x^3 - x^2 + 8$ .

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