Question

Factor and find all solutions of x^4 - 4x^3 - x^2 +16x = 12

Answer 1

\[x^4-4x^3-x^2+16x=12\]

Answer 2

I just don't know how to start it...

Answer 3

Start with: \[
x^4-4x^3-x^2+16x-12=0
\]

Answer 4

Yea but then i can't take out the GCF

Answer 5

Because it was not a GCF to begin with.

Answer 6

But that won't help you find roots.

Answer 7

That's why i'm confused lol

Answer 8

One think we can try is splitting the \(-x^2\) term.

Answer 9

I was thinking maybe you could group them...

Answer 10

But then how would you bring the 12 into it

Answer 11

hmm

Answer 12

(By the way, I'm going of intuition here, this might not work, but it's an attempt)

Answer 13

u can guess a bunch but a nice way to see where ur zeros lie is to take first and 2nd derivatives

Answer 14

and graph and see where ur zeros should be

Answer 15

I was never really taught how to graph these...

Answer 16

they give u nice ones, try guessing for this one though, 1,-1, 2,-2, 4,-4...

Answer 17

Notice: \[
\color{red}{1}x^4\color{blue}{-4}x^2 +ax^2+\color{red}{b}x^2\color{blue}{+16}x-12
\]We want: \[
\frac{1}{-4} = \frac{b}{16}
\]Which would imply that \(b=-4\).

Answer 18

Now since \(a+b=-1\) and \(b=-4\), that means \(a+-4 = -1\implies a=3\).

Answer 19

Where did u get the extra -x^4 from?

Answer 20

This gives us: \[
(x^4-4x^3+3x^2)+(-4x^2+16x-12)
\]

Answer 21

What you could do is a binomial times a trinomial, no?

Answer 22

Now, we can pull out the GFCs: \[
x^2(x^2-4x+3)+-4(x^2-4x+3)
\]

Answer 23

Which gives us: \[
(x^2-4)(x^2-4x+3)
\]

Answer 24

This method is called "splitting the middle"

Answer 25

\[(x^4-4x^3)(-x^2+16x-12)\]
Could u just do that from the very first equation?

Answer 26

No, you can't. They are being added, not multiplied.

Answer 27

oh

Answer 28

Besides, we aren't completely done yet.

Answer 29

\[
x^2-4 = x^2-2^2 = (x-2)(x+2)
\]This is called difference of squares.

Answer 30

\[
x^2-4x+3 = (x^2-x) + (-3x+3) = x(x-1)+-3(x-1) = (x-3)(x-1)
\]Splitting the middle another time.

Answer 31

Sry i gtg i'll just try and ask my friend tommorrow morning u were a great help i just ran out of time x.x

Answer 32

Final solution:\[
(x-2)(x+2)(x-3)(x-1)=0
\]