# How do I factor something with 3 parts with a higher power than 2 and a coefficient in (a)?

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.

• You can google "how to factor a trinomial" to see explanations and videos. – Browning Aug 13 '16 at 22:08
• The short answer is that you should look for "easy factors." By fundamental theorem of algebra, for a polynomial $f(x)$, if $f(a)=0$ then $a$ is a "root" of the polynomial, implying that $f(x)=(x-a)\cdot g(x)$ where $g(x)$ is another polynomial of smaller degree than $f$. The easiest roots to check for are $0,1,-1$. If you can manage to factor out some of the simpler roots and you are left with a quadratic, you can use the well known quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the remaining roots. Larger degree polynomials can be trickier (and might not even be possible). – JMoravitz Aug 13 '16 at 22:13
• Closed form solutions exist for finding the roots of polynomials with degree at most four, the formula for a quadratic almost being common knowledge. The formula for finding the roots of a general degree three polynomial is much less common to know but exists. The formula for general fourth degree polynomials would take several pages to write down completely. It is a famous result of Galois Theory that there does not and can not exist a closed form solution for general degree 5 or higher polynomials. – JMoravitz Aug 13 '16 at 22:16
• Welcome to the site. If you put dollars around the math it'll render nicely. If you want to do more elaborate things later see meta.math.stackexchange.com/questions/5020/… – quid Aug 13 '16 at 22:16
• @Ed_4434 of course. $f(x)=x^5$ being an easy example of one which can be factored: $x^5=(x-0)(x-0)(x-0)(x-0)(x-0)$, but my point being, if you give an arbitrary polynomial $ax^5+bx^4+cx^3+dx^2+ex+f$ there is no easy solution. Even given specific numbers like $x^5+x^4-3x^3+8x^2-x+5$, although we know that a factoring must exist, the factors themselves cannot be easily expressed. If it happens to have one, it would be a special case and would largely be a fluke. – JMoravitz Aug 13 '16 at 22:25

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.