Question

h

Answer 1

First, can you factor the equation at all?

Answer 2

\[ f_{x} = x^4 - 4x^3+3x^2\]

What's similar about each term?

Answer 3

Correct! But each of them have a power of at least \[x^2\]

so you can pull them out front
\[f_{x} = x^2(x^2 - 4x+3)\]

Answer 4

Yeah, exactly. So you'll have 4 solutions for x.
x = 0
x = 0
from the x^2

Then two solutions from factoring the polynomial in the parenthesis

Answer 5

Yup. Either b or d

Answer 6

Then
\[x^2 - 4x + 3\]
can be factored into

\[(x-3)(x-1)\]

Answer 7

set those both equal to zero and solve for x

Answer 8

nope. The zeroes are for when you solve the equation. as in "If x were this number, this term would be zero"

if x = -3, then in your expression

(-3 -3) = -6 ≠ 0

So for

(x-3) = 0
x = 3

Answer 9

So B

Answer 10

sorry for the tardiness :/