Factor 32x^2 y^2 - 24x.

Step-by-step solution

1. 1

   Find the numerical part of the GCF. $32=2^5$ and $24=2^3\cdot 3$, so the largest common numerical factor is $2^3=8$.

2. 2

   Find the variable part of the GCF. The first term has $x^2y^2$, the second has only $x$. The shared power of $x$ is the smaller exponent, $x^1$; and since $y$ is absent from the second term, no $y$ can be taken out:

   $\text{GCF}=8x$

3. 3

   Divide each term by the GCF.

   $\frac{32x^2y^2}{8x}=4xy^2,\qquad\frac{-24x}{8x}=-3$

   $32x^2y^2-24x=8x\left(4xy^2-3\right)$

4. 4

   Check whether the bracket factors further. $4xy^2-3$ is not a difference of squares ($3$ is not a perfect square and $4xy^2$ is not a square because of the lone $x$), and its two terms share no common factor. So the factorisation is complete.

5. 5

   Verify by expanding. $8x\cdot 4xy^2=32x^2y^2$ and $8x\cdot(-3)=-24x$ $\checkmark$. Numerically at $x=1$, $y=2$: original $=128-24=104$, and $8(16-3)=8\cdot 13=104$ $\checkmark$.