Simplify (6x − 6) / (12x² + x(x² − 2x + 1)).

Step-by-step solution

1. 1

   Factor the numerator. Both terms share a factor of $6$:

   $6x-6=6(x-1)$

   The visible $(x-1)$ is what makes it tempting to expect a cancellation — the rest of the work decides whether one actually exists.

2. 2

   Take out the common x in the denominator. The perfect square $x^2-2x+1=(x-1)^2$, so

   $12x^{2}+x(x-1)^{2}=x\left[12x+(x-1)^{2}\right]$

   Note that $x$ — not $x^2$ — is the highest common factor, because the second term carries only one $x$.

3. 3

   Expand and collect inside the bracket. This is the step that decides the answer:

   $12x+(x-1)^{2}=12x+x^{2}-2x+1=x^{2}+10x+1$

   The $(x-1)^2$ does not survive as a factor once it is added to $12x$.

4. 4

   Assemble and test for a common factor.

   $\frac{6(x-1)}{x\left(x^{2}+10x+1\right)}$

   Would $(x-1)$ divide $x^2+10x+1$? Evaluate at $x=1$: $1+10+1=12\neq 0$, so by the factor theorem it does not. Nothing cancels.

5. 5

   State the simplified form.

   $\boxed{\dfrac{6(x-1)}{x\left(x^{2}+10x+1\right)}}$

   valid for $x\neq 0$ and $x\neq -5\pm 2\sqrt6$ (the roots of $x^2+10x+1$).

6. 6

   Spot-check numerically. At $x=3$ the original is $\dfrac{12}{108+3(4)}=\dfrac{12}{120}=0.1$, and the simplified form gives $\dfrac{6(2)}{3(9+30+1)}=\dfrac{12}{120}=0.1$. They agree, so the factoring introduced no error.