# factorizing involves xy yz zx terms

• May 17th 2008, 07:58 AM
afeasfaerw23231233
factorizing involves xy yz zx terms
p7 q13
-question: factorize $x^2-3y^2-8z^2+2xy+14yz+2zx$

my working seems too long. is there any better method?
my working:
let $(x+ay+bz)(x+cy+dz)$
expand it: $x^2+(a+c)xy+(b+d)xz+acy^2+(ad+bc)yz+bdz^2$
equivalent the coefficient:
ac=-3
a+c=2
a(2-a)=-3
a^2-2a-3=0
a=3 or -1
c=-1 or 3
same on b and d
when $(a, c) = (3, -1), 3d-b = 14$
$(d,b)=(4 , -2)$
when $(a,c) = (-1,3), (d,b) = (-2, 4)$
so $(x+3y-2z)(x-y+4z)$
too long! is there any better method? thanks
• May 17th 2008, 08:27 AM
Moo
Hello,

Quote:

Originally Posted by afeasfaerw23231233
p7 q13
-question: factorize $x^2-3y^2-8z^2+2xy+14yz+2zx$

my working seems too long. is there any better method?
my working:
let $(x+ay+bz)(x+cy+dz)$
expand it: $x^2+(a+c)xy+(b+d)xz+acy^2+(ad+bc)yz+bdz^2$
equivalent the coefficient:
ac=-3
a+c=2
a(2-a)=-3
a^2-2a-3=0
a=3 or -1
c=-1 or 3
same on b and d
when $(a, c) = (3, -1), 3d-b = 14$
$(d,b)=(4 , -2)$
when $(a,c) = (-1,3), (d,b) = (-2, 4)$
so $(x+3y-2z)(x-y+4z)$
too long! is there any better method? thanks

I know a method, but I don't know if it's for your level...

Group the terms containing x.

${\color{red}x^2+2xy+2xz}-3y^2-8z^2+14yz={\color{red}x^2+2x(y+z)}-3y^2-8z^2+14yz$

Complete the square :

$x^2+2x(y+z)=(x+y+z)^2{-\color{blue}(y+z)^2}$

-->
\begin{aligned} f(x) &=(x+y+z)^2{\color{blue}-y^2-z^2-2yz}-3y^2-8z^2+14yz \\
&=(x+y+z)^2-4y^2+12yz-9z^2 \\
&=(x+y+z)^2-\left((2y)^2-2 \cdot (2y) \cdot (3z)+(3z)^2\right) \\
&=(x+y+z)^2-(2y-3z)^2 \\
&=\left(x+y+z-(2y-3z)\right)\left(x+y+z+(2y-3z)\right) \\
&=(x-y+4z)(x+3y-2z) \end{aligned}

Does it make sense?