Factor the expression (a + x)^(m+1) (b + x)^(n-1) - (a + x)^m (b + x)^n completely.

Step-by-step solution

1. 1

   Read the exponents, not the letters. Both terms are built from the same two bases, $(a+x)$ and $(b+x)$; only the exponents differ. Term 1 carries $(a+x)^{m+1}(b+x)^{n-1}$ and term 2 carries $(a+x)^{m}(b+x)^{n}$. When two products share bases, the greatest common factor uses the smaller exponent of each base — that rule is the whole method here, and it works even though $m$ and $n$ are unknown.

2. 2

   Take the smaller exponent of each base. For $(a+x)$ the exponents are $m+1$ and $m$, so the smaller is $m$. For $(b+x)$ they are $n-1$ and $n$, so the smaller is $n-1$. Hence

   $\text{GCF}=(a+x)^{m}(b+x)^{n-1}$

   Note that this is legitimate for any integers $m,n$: we never had to know their values, only which of each pair is smaller.

3. 3

   Divide each term by the GCF using $x^p/x^q=x^{p-q}$.

   $\frac{(a+x)^{m+1}(b+x)^{n-1}}{(a+x)^{m}(b+x)^{n-1}}=(a+x)^{1}=a+x$

   $\frac{(a+x)^{m}(b+x)^{n}}{(a+x)^{m}(b+x)^{n-1}}=(b+x)^{1}=b+x$

   So the expression becomes $(a+x)^m(b+x)^{n-1}\left[(a+x)-(b+x)\right]$.

4. 4

   Simplify the bracket — this is where the problem collapses. Distribute the minus sign carefully:

   $(a+x)-(b+x)=a+x-b-x=a-b$

   The two $x$ terms cancel, so the bracket is not a binomial in $x$ at all; it is the constant difference $a-b$. Forgetting to distribute the minus over $+x$ is the single most common error and would leave a wrong bracket $a-b+2x$.

5. 5

   Write the factored form and check it. Putting the pieces together,

   $(a-b)(a+x)^{m}(b+x)^{n-1}$

   Quick numeric check with $a=2,b=1,x=1,m=2,n=3$: the original is $3^{3}\cdot 2^{2}-3^{2}\cdot 2^{3}=108-72=36$, and the factored form gives $(2-1)\cdot 3^{2}\cdot 2^{2}=36$. The two agree, so the factorisation is correct.