I am running into problems simplifying using factoring of binomial expressions. The problem at hand is this:
$(x-1)^3*(2x-3)-(2x+12)*(x-1)^2$
I first expanded the left side of the minus sign, like so:
$(x-1)(x-1)(x-1)(2x-3)$
and multiplied the factors together to get:
$2 x^4-9 x^3+15 x^2-11 x+3$
I did this so I can combine the terms with those on the right side of the minus sign. So I continued with the right side of the minus sign:
$2 x^3+8 x^2-22 x+12$
and combined both sides to get:
$2 x^4-11 x^3+7 x^2+11 x-9$
I then proceeded to try and factor this (as the instructions note the answer must be in factored form). Failing at factoring here, I turned to Wolfram Alpha and input the expression. This was the result:
$(x-1)^2 (x+1) (2 x-9)$
The same result is given when I input the original expression.
Did I go about doing this problem in a ill-advised way? Is there a simpler way to do this without running into incredibly difficult expressions that need to be factored? If not, is there an appropriate way to factor expressions like the ones listed above?
$\endgroup$2 Answers
$\begingroup$There is indeed a simpler way to do it!
Always look for common factors at the very beginning. In this case, we can factor out $(x-1)^2$ to give $$(x-1)^3(2x-3)-(2x+12)(x-1)^2 \equiv (x-1)^2[(x-1)(2x-3)-(2x+12)].$$
Now all we have to do is expand quadratic terms which is much easier!
We then get
$$(x-1)^2[2x^2-7x-9] \equiv \boxed{(x-1)^2(2x-9)(x+1)}$$
$\endgroup$ 2 $\begingroup$From the very first line you can see that $(x-1)^2$ is a common factor, i.e.
$$ (x-1)^2[(x-1)(2x-3) - (2x+12)] $$
expanding only that in the square bracket
$$ (x-1)^2[2x^2-7x-9] $$
And then factorise the quadratic to get
$$ (x-1)^2(x+1)(2x-9) $$
$\endgroup$
