How do you fully factor $4 x^{2} - 6 x y$ $4x^2-6xy$ ?

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Answer 1 · 2018-03-19T09:22:37

$2 x (2 x - 3 y)$ $2x(2x-3y)$

we take out the hcf from each term
so

$h c f (4, 6) = 2$ $hcf(4,6)=2$

$h c f (x^{2}, x) = x$ $hcf(x^2,x)=x$

$:. 4x^2-6xy=2x(2x-3y)$

Answer 2 · 2018-03-19T09:24:17

$4x^2-6xy=2x(2x-3y)$

Factor out the common term. Here the common term is 2x.

Thus,

$4x^2-6xy=2x(2x-3y)$

How do you fully factor 4x^2-6xy4x2−6xy4x^2-6xy?