Multiply and simplify: (x² − 2x + 1)/(x² + 2x − 3) × (x² + 3x)/(x² + 2x).

Step-by-step solution

1. 1

   Factor everything before multiplying. Multiplying the numerators and denominators out first would produce two quartics that then have to be re-factored. Factoring up front lets the cancellations happen while the pieces are still small.

2. 2

   Factor the four quadratics.

   $x^{2}-2x+1=(x-1)^{2},\qquad x^{2}+2x-3=(x+3)(x-1)$

   $x^{2}+3x=x(x+3),\qquad x^{2}+2x=x(x+2)$

3. 3

   Rewrite the product in factored form.

   $\frac{(x-1)^{2}}{(x+3)(x-1)}\cdot\frac{x(x+3)}{x(x+2)}$

4. 4

   Cancel across both fractions. In a product, any numerator factor may cancel with any denominator factor. One $(x-1)$ cancels, $(x+3)$ cancels, and $x$ cancels:

   $\frac{x-1}{1}\cdot\frac{1}{x+2}=\frac{x-1}{x+2}$

   $\boxed{\dfrac{x-1}{x+2}}$

5. 5

   Record the excluded values. The cancellations are only legal where the cancelled factors are nonzero, so the identity holds for $x\neq 0$, $x\neq 1$, $x\neq -3$ and $x\neq -2$. The final form only shows the restriction $x\neq -2$; the other three must be carried along from the original.

6. 6

   Check with a test value. At $x=4$: the first fraction is $\tfrac{9}{21}=\tfrac37$ and the second is $\tfrac{28}{24}=\tfrac76$, whose product is $\tfrac{3}{6}=\tfrac12$. The simplified form gives $\tfrac{4-1}{4+2}=\tfrac36=\tfrac12$. They match.