Question

Ok i'm having problems factoring
3x^2-4x+2

Answer 1

\[3x^2-4x+2\]

Answer 2

use the quadratic equation?

Answer 3

3*2 = 6

--> no factors of 6 add up to -4
Does not factor

Answer 4

alright i guess i will use the quadratic, I just wanted to make sure their wasn't another pattern i was forgetting.

Answer 5

\[b=(\frac{-4}{2}+z)+(\frac{-4}{2}-z)\]
\[3(2)=6=ac=(\frac{-4}{2}+z)(\frac{-4}{2}-z)\]

we need to solve
\[6=(\frac{-4}{2}+z)(\frac{-4}{2}-z)=(\frac{-4}{2})^2-z^2\]
so we have
\[6=4-z^2=> z^2=4-6=-2 => z^2=-2 => z=\pm \sqrt{-2}=\pm i \sqrt{2}\]

we will just be needing \[z=i \sqrt{2}\]

so bx=\[(-2+i \sqrt{2})x+(-2-i \sqrt{2})x\]

so we replace the bx term in ax^2+bx+c with the above

\[3x^2+(-2+i \sqrt{2})x+(-2- i \sqrt{2})x+2\]

now factor by grouping!


\[3x(x+\frac{-2+i \sqrt{2}}{3})+(-2 - i \sqrt{2})(x+\frac{2}{-2- i \sqrt{2}})\]

now we want to see if \[\frac{-2 + i \sqrt{2}}{3} and \frac{2}{-2- i \sqrt{2}}\]
to do this we should rationalize the denominator of this second one

\[\frac{2}{-2-i \sqrt{2}} \cdot \frac{-2+i \sqrt{2}}{-2+i \sqrt{2}}=\frac{2(-2+i \sqrt{2})}{4-i^2 (2)}=\frac{2(-2+i \sqrt{2})}{4-(-1)(2)}\]
\[=\frac{2(-2+i \sqrt{2})}{6}=\frac{-2+ i \sqrt{2}}{3}\]

so we have

\[3x(x+\frac{-2+i \sqrt{2}}{3})+(-2 - i \sqrt{2})(x+\frac{-2+i \sqrt{2}}{3})\]

so each term has the factor \[x+\frac{-2 + i \sqrt{2}}{3}\] in common we can factor that out
\[(x+\frac{-2+i \sqrt{2}}{3})(3x-2-i \sqrt{2})\]

Answer 6

This does not have real solutions ... its solutions are complex

Answer 7

these being the solutions
x = 1/3 (2-i sqrt(2))
x = 1/3 (2+i sqrt(2))

Answer 8

Depending what level of math you are at. It cannot be factored unless you use imaginary numbers and you might not of be at that level yet. So it cannot be factored

Answer 9

I'm getting

\[\frac{4\pm2i \sqrt2}{6}\]

can you show me where i'm going wrong?

Answer 10

you can reduce but looks good! :)

Answer 11

that is correct, just needs to be simplified, factor out a 2

Answer 12

oh doh! thanks guys