Question

Factorize \(x^2 - 3y^2 - 8z^2 + 2xy + 14yz + 2zx\)

Answer 1

rearrange
x^2 + 2xy - 3y^2 -8z^2 + 2zx + 14yz
= (x - y)(x + 3y) - 2z(4z - 7t + x)

Answer 2

That doesn't look nice :( How to continue?

Answer 3

yes you are right that should be 7y not 7t
but other than that this does not look like the way to go

Answer 4

this probably will factor into 2 trinomials..

Answer 5

\[
x^2+2 x y+2 x z-3 y^2+14 y z-8 z^2=(x+3 y-2 z) (x-y+4 z)
\]

Answer 6

I'm actually thinking this, but it will be EXTREMELY UGLY!
\[(A_1x+B_1y+C_1z)(A_2x+B_2y+C_2z)\]\[=A_1A_2x^2+B_1B_2y^2+C_1C_2z^2\]\[+(A_1B_2+A_2B_1)xy+(B_1C_2+B_2C_1)yz+(A_1C_2+C_1A_2)xz\]
And solve those A1, A2, ...C1, C2..

Answer 7

@eliassaab Would you mind showing how you did it?

Answer 8

Apply your method. You know for sure that A1=A2=1
You can also do some guessing about C1 and C2 since C1C2=-8
and so fourth.

Answer 9

Also B1 B2=-3

Answer 10

aah !

Answer 11

Oh.. I'm stupid! :(
Thanks!!

Answer 12

no .. you're not ..

Answer 13

So,
A1 = 1
A2 = 1

(Since C1C2 = -8 and C1+C2 = 2)
C1 = -2
C2 = 4

When B1 =\(\pm\) 1 or B1 = \(\pm\)3
Then B2 = \(\mp\) 3 or B1 = \(\mp\)1
When B1 = 1 and B2 = -3
B1C2 + C1B2
4 + 6 = 10 (rejected)
When B1 =-1 and B2 = 3
B1C2 + C1B2
= -4 - 6
= -10 (rejected)
When B1 = 3 and B2 = -1
B1C2 + C1B2
12 + 2 = 14

So, B1 = 3 and B2 = -1
Therefore,
x^2 - 3y^2 - 8z^2 + 2xy + 14yz + 2zx
= (x+3y-2z)(x-y+4z)

Wow!
Thanks again!!

Answer 14

yes ..