Question

-x^5-x^3+6 how do you factor this?

Answer 1

i think it might be easier to factor
\[-(x^5+x^3-6)\]

Answer 2

this is what I got -x(x^4+x^2+6). is it right?

Answer 3

or maybe not
not clear how to factor this at all

Answer 4

no, not unless you had a typo in the problem

Answer 5

-x^5-x^3+6x
oops yea sorry it had 6x

Answer 6

could it have been
\[-x^5-x^3+6x\]?

Answer 7

yes, I messed up

Answer 8

ok i see
your answer was close, but not quite
should have been
\[ -x(x^4+x^2-6)\]

Answer 9

then you can continue to factor, since \(x^4+x^2-6\) factors

Answer 10

I'm didn't know you could keep factoring..?
I'm trying to find the x intercept afterwards, could you help me walk through this one?

Answer 11

sure
if it was \(x^2+x-6\) you could probably factor it easy right?

Answer 12

right that would have been (x-2)(x+3)

Answer 13

right, and this is exactly the same, except instead of \(x^2\) and \(x\) you have \(x^4\) and \(x^2\)
noting that \(x^4=(x^2)^2\)

Answer 14

so it factors in exactly the same way, but with \(x^2\) instead of \(x\) i.e.
\[x^4+x^2-6=(x^2-2)(x^2+3)\]

Answer 15

final answer therefore looks like
\[-x(x^2-2)(x^2+3)\]

Answer 16

oh, so you would ignore the -x on the outside first and factor what's in the parenthesis. And to find the x-int. you would set y=o right?
f(x)=-x(x^2-2)(x^2+3)
0=-x(x^2-2)(x^2+3)
x=2 x=-3 ?

Answer 17

oh no

Answer 18

\[0=-x(x^2-2)(x^2+3)\] is right

Answer 19

then \(-x=0\iff x=0\) is also right, but
\[x^2-2=0\] is not \(x=2\)

Answer 20

\[x^2-2=0\\
x^2=2\\
x=\pm\sqrt2\] is more like it

Answer 21

\[x ^{2}+3=x=\pm \sqrt{-3}=\pm i \sqrt{3}?\]

Answer 22

how would you graph these as a sq root?