# Thread: factorizing involves xy yz zx terms

1. ## factorizing involves xy yz zx terms

p7 q13
-question: factorize $\displaystyle x^2-3y^2-8z^2+2xy+14yz+2zx$

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

2. Hello,

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

my working seems too long. is there any better method?
my working:
let $\displaystyle (x+ay+bz)(x+cy+dz)$
expand it: $\displaystyle x^2+(a+c)xy+(b+d)xz+acy^2+(ad+bc)yz+bdz^2$
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 $\displaystyle (a, c) = (3, -1), 3d-b = 14$
$\displaystyle (d,b)=(4 , -2)$
when $\displaystyle (a,c) = (-1,3), (d,b) = (-2, 4)$
so $\displaystyle (x+3y-2z)(x-y+4z)$
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.

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

Complete the square :

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

-->
\displaystyle \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?