Question

x^5-12x^3+32x=0
Please help and show work

Answer 1

x(x^4-12x^2+32)=0
x((4+-1x)(8+-1x))=0

Answer 2

x^5-12x^3+32x=0
x(x^4-12x^2+32)=0
x^4-12x^2+32=0

let x^2=a

a^2-12a+32=0

Do it from here.

Answer 3

Can you continue from there?

Answer 4

how @SolomonZelman get X^2=a?

Answer 5

I didn't get it, I am substituting a for x^2 so that it would be easier to solve.

it's a make up.

Answer 6

so either x=0, 4+-1x=0, or 8+-1x=0. so x {0,4,8}

Answer 7

you didn't show any work, you do realize that.

Answer 8

i'm confused

Answer 9

x^5-12x^3+32x=0 factor out of x,
x(x^4-12x^2+32)=0 divide both sides by x,
x^4-12x^2+32=0

let x^2=a (substituting "a" for "x^2" )

it becomes,
a^2-12a+32=0

can you solve this quadratic?

Answer 10

Or you can forget about a=x^2, just say

x^4-12x^2+32=0
(x^2-4)(x^2-8)=0

x^2-4=0 or x^2-8=0
x^2=4 or x^2=8\
x= +- 2 or x= +- 2 sqrt(2)

Answer 11

I subtract 32 from zero first right

Answer 12

oh so there are two answers?

Answer 13

@SolomonZelman

Answer 14

I can't start from a scratch.

Answer 15

huh?