Factorise 3x^2+4y^2+25z^2-4√3xy-20yz+10√3zx.

Question

This is from chapter polynomial.

Answer 1

Note that there is no constant term so assume:

(ax+by+cz)(dx+ey+fz)=adx^2+bey^2+cfz^2+(ae+bd)xy+(cd+af)xz+(bf+ce)yz.

ae+bd=-4√3; cd+af=10√3; bf+ce=-20; ad=3; be=4; cf=25.

If a=d=√3 (because ad=3), then e√3+b√3=-4√3, so e+b=-4, and since be=4, b=e=-2.

c√3+f√3=10√3 and c+f=10, and, since cf=25, c=f=5.

bf+ce=-10-10=-20, so we have the set! Factorisation is: (x√3-2y+5z)(x√3-2y+5z) = (x√3-2y+5z)^2.

Answer 2

(3x−4y)^2−25z^2
=(3x−4y^)2−(5z)^2, Using a^2−b^2=(a+b)(a−b)
=(3x−4y+5z)(3x−4y−5z)