I'm starting college this year, and I had to take a year off in order to collect some funds to help pay for it. I really don't remember how to factor something such as
$4x^2 + 4x - 35$
I know the answer to the problem to this question since it's a part of my readiness quiz study guide. But it was done in 1 step.
The answer to it is (2x-5) (2x+7). No other steps were done to factor the problem, it was just done in one step unlike the others such as $5x^3 - 20x$ which I understand how to do. And I understand how to do something like
$x^3 + 2x^2 - 9x - 18$. I can't really explain the terms used to solve something like this, but I get how to do it.
Another example of a problem I can't solve due to the rules of the formula is
$8x^4-17x^3+9x^2$
That ends up just becoming $x^2 (8x^2 - 17x + 9)$ but where do I go from there? I have to factor the same way just like in the $4x^2 + 4x - 35$ problem.
Can someone please explain how this is supposed to be approached. I'd really appreciate it. Thank you.
I'll go over a known method which I know as the $AC$-method and will use your polynomial, $$4x^2+4x-35.$$
The method is called the $AC$-method because assuming the polynomial is of the form $Ax^2+Bx+C$, we want to look at $AC$. For our polynomial, $A=4, B=4, C=-35$ and we see $AC= (4)(-35) = -140$.
Next, we identify two integers $X,Y$ such that their product is $AC= -140$ and their sum is $B = 4$. After exploring factor pairs of $-140$, we see that $X=14, Y= -10$ works since $XY = (14)(-10)=-140=AC$ and $X+Y = 14+(-10) = 4=B$. Now we use our $X$ and $Y$ to break up the middle term. From there, we group terms and factor, $$4x^2+4x-35$$ $$=4x^2+14x-10x-35$$ $$=(4x^2+14x)+(-10x-35)$$ $$= 2x(2x+7)-5(2x+7)$$
Notice both terms have a factor of $2x+7$ we can factor out, $$2x(2x+7)-5(2x+7)$$ $$= (2x+7)(2x-5).$$
This method will work whenever your quadratic is factorable over the rationals, i.e., it can be written as $(ax+b)(cx+d)$ where $a,b,c,d$ are integers.