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Math Help - factorizing involves xy yz zx terms

  1. #1
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    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
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    Hello,

    Quote Originally Posted by afeasfaerw23231233 View Post
    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 \\<br />
&=(x+y+z)^2-4y^2+12yz-9z^2 \\<br />
&=(x+y+z)^2-\left((2y)^2-2 \cdot (2y) \cdot (3z)+(3z)^2\right) \\<br />
&=(x+y+z)^2-(2y-3z)^2 \\<br />
&=\left(x+y+z-(2y-3z)\right)\left(x+y+z+(2y-3z)\right) \\<br />
&=(x-y+4z)(x+3y-2z) \end{aligned}


    Does it make sense?
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