# Universal factoring method or list of methods (trinomials)

I am a student in calculus II. I'm now failing tests solely because I cannot factor; I understand everything else. This is compounded by the fact it seems to exceedingly hard to find anything comprehensive online. This is a long-running problem I've been facing; I can't be the only one.

I know some methods, but I always run into more problems I can't factor. I know about synthetic/long division, binomial/perfect square trinomial factoring, difference and sum of cubes and squares, the grouping method, the quadratic formula (unsure how to use for factoring), guess and check, and some trinomial methods. I only know the difference/sum of squares/cubes well. For example, I can't factor $4x^2 - 4x - 3$.

It would be dandy if I could get a comprehensive list or a universal solution to all these factoring questions, but that's unlikely to happen (and would be too broad). I'll narrow this down to something more tangible: What are the methods I should know to factor trinomials?

For factoring simple quadratic equations, it's simply a matter of remembering simple forms and tricks. For example, one of the easiest quadratic factorings is the difference of squares

$a^2 -b^2 = (a+b)(a-b)$

Here are several examples of difference of squares factoring:

$x^2 - 1 = (x+1)(x-1) \\ 9x^2 -4 = (3x+2)(3x-2) \\ 36x^2 - 25y^2 = (6x +5y)(6x-5y)$

The next simplest factoring is quadratics without constant terms, which boils down to just spotting common factors. Ex: $9x^2 +3x = 3x(3x+1)$

Next there is factoring quadratics with a leading coefficient of 1, which amounts to pairing factors:

If $x^2 + hx +k = (x+a)(x+b)$, then $k = ab$ and $h = a+b$. This comes from a simple expansion of $(x+a)(x+b) = x^2 + ax + bx + ab = x^2 + (a+b)x + ab$

For example, $x^2 + 20x + 36$. The factors of $36$ are $(1,36) \ (2,18) \ (3,12) \ (4,9) \ (6,6)$ Notice that $20 = 2+18$. Thus, $x^2 + 20x + 26 = (x+2)(x+18)$. This can always be checked by re-expanding.

This even works with negative terms. For example, $x^2 - 4x -12$ The factors of $-12$ are $(-1, 12) \ (-2,6) \ (-3, 4) \ (-4,3) \ (-6, 2) \ (-12, 1)$. Notice that $-6 + 2 = -4$. Thus, $x^2 - 4x -12 = (x-6)(x+2)$

The last case is when the leading coefficient is anything other than $1$. This can be done with a method similar to the one above, but with slightly more computation. For any quadratic polynomial $ax^2 + bx + c$, if it has an expansion in integers $(px + q)(rx +s)$, then $a = pr$, $c = qs$, and $b = (ps + qr)$. Thus, this method of factoring involves finding all the factors of $a$ and $c$, and pairing them in such a way as to equal the middle term $b$. It's easier to demonstrate than to explain.

Let's use your example: $4x^2 - 4x - 3$. The factors of $4$ are $(1,4) \ (2,2)$ and the factors of $-3$ are $(-1,3) \ (1, -3)$. Let's begin pairing exhaustively.

$(1)(-1) + (4)(3) = 11 \\ (1)(3) + (4)(-1) = -1 \\ (1)(1) + (4)(-3) = -11 \\ (1)(-3) + (4)(1) = 1 \\ (2)(-1) + (2)(3) = 4 \\ (2)(1) + (2)(-3) = -4$

Every one of these expressions is equivalent to (a factor of $a$)(a factor of $b$) + (the corresponding factor of $a$)(the corresponding factor of $b$)

Notice that the last expression is the one we need. We need to take the numbers $2, 1, 2, -3$ and plug them into $(\_x +\_)(\_x+\_)$ such that the product is equal to $4x^2 -4x -3$

We know the coefficients of the $x$ terms will be $(2x+\_)(2x+\_)$, because they must multiply to be $4$. From here, we would place the $1, -3$ in the expression such that they match up with the appropriate term to multiply by. However in this case since it's just two $2$'s, the answer is $4x^2 - 4x -3 = (2x+1)(2x-3)$. Check this by expanding back out.

Another example of this method: $3x^2 -7x + 2$. The factors of $3$ are $(1,3)$ and the factors of $2$ are $(1,2)$. Notice that the middle term is negative, though, so we must use $(-1, -2)$ as our factors of $2$. Now we exhaustively pair.

$(1)(-2) + (3)(-1) = -5 \\ (1)(-1) + (3)(-2) = -7$.

The last expression is the one we need. Thus, the factored expansion is $(x+\_)(3x+\_)$. Notice here that for the expansion to be correct, the $3$ must multiply by $-2$. Thus, we plug in into the factor $(x+\_)$, otherwise $-2$ and $3$ would not be multiplied.

Thus, $3x^2 - 7x + 2 = (x-2)(3x-1)$. Check this by expanding back out.

If the method of factor-pairing by exhaustion reveals that there is no possible combination of factors which will sum to the middle term, then the expression has no factorization in integers.

• Thank you for the comprehensive list. That really helped me organize my thoughts. I'm building a little compendium of what I know and this provided the spark to help me understand how roots provide factors using the F.T.A., which easily explains why the quadratic formula is useful in factoring. – person27 Mar 18 '16 at 3:34

Before we consider how to factor $4x^2 - 4x - 3$, let's see what happens when we form the product of two linear monomials.

\begin{align*} (6x - 7)(3x + 5) & = 6x(3x + 5) - 7(3x + 5)\\ & = \color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}\\ & = \color{blue}{18}x^2 - \color{green}{9}x - \color{blue}{35} \end{align*} To factor, what we wish to do is carry out the steps in reverse. The key observation is that in the expression $\color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}$, the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is the linear coefficient of the product. That is, $$(\color{blue}{18})(\color{blue}{-35}) = (\color{green}{30})(\color{green}{-21}) = -630$$ This suggests that we can factor a quadratic polynomial $ax^2 + bx + c$ with rational coefficients with respect to the rationals if we can split the linear term if we can find two numbers with product $ac$ and sum $b$.

In your example of $4x^2 - 4x - 3$, we can split the linear term if we can find two numbers with product $(\color{blue}{4})(\color{blue}{-3}) = -12$ and sum $-4$. The possible factorizations of $-12$ are \begin{align*} -12 & = -1 \cdot 12 & -12 & = 1 \cdot -12\\ & = -2 \cdot 6 & & = \color{green}{2} \cdot \color{green}{-6}\\ & = -3 \cdot 4 & & = 3 \cdot -4 \end{align*} By inspection, the only pair of factors of $-12$ with sum $-4$ are $2$ and $-6$. Hence, \begin{align*} 4x^2 - 4x - 3 & = 4x^2 + 2x - 6x - 3 && \text{split the linear term}\\ & = 2x(2x + 1) - 3(2x + 1) && \text{factor by grouping}\\ & = (2x - 3)(2x + 1) && \text{extract the common factor} \end{align*} Another option is completing the square. \begin{align*} 4x^2 - 4x - 3 & = 4(x^2 - x) - 3\\ & = 4\left(x^2 - x + \frac{1}{4}\right) - 1 - 3\\ & = 4\left(x - \frac{1}{2}\right)^2 - 4\\ & = 4\left[\left(x - \frac{1}{2}\right)^2 - 1\right]\\ & = 4\left[\left(x - \frac{1}{2}\right) + 1\right]\left[\left(x - \frac{1}{2}\right) - 1\right]\\ & = 4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right) \end{align*} Since \begin{align*} 4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right) & = 2 \cdot 2 \left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right)\\ & = \left[2\left(x + \frac{1}{2}\right)\right]\left[2\left(x - \frac{3}{2}\right)\right]\\ & = (2x + 1)(2x - 3) \end{align*} our two factorizations are equivalent.

For the general case, \begin{align*} ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x\right) + c\\ & = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c\\ & = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}\\ & = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right]\\ & = a\left[x + \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right]\left[x + \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right] \end{align*}

• Ah, so that's what I did on that particular example that threw me off. I somehow forgot the 4 and left it as $(x + 1/2)(x - 3/2)$, then scratched my head looking at the answer and wondering what I did wrong. Good to know! – person27 Mar 18 '16 at 3:03

(I know mostly link answers are discouraged, but I don't have much time now)

For factoring trinomials such as $4x^2 - 4x - 3,$ one method is what I guess could be called "the quadratic formula method", which I describe in this 11 November 2008 ap-calculus post archived at Math Forum, and another method is the "AC Method", which is described in the math education stackexchange question Factoring quadratics where the coefficient on the $x^2$ term does not equal 1.

For polynomials in general, you can use the factor theorem, which I explain by examples in my answer to Finding limit of a quotient and which you can practice with using this old worksheet of mine.

Firstly, you can always check if you factored correctly using the quadratic formula,

$$ax^2+bx+c=a(x-r_1)(x-r_2)$$

Where $$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

But of course, I imagine your aim is to factor it when all of the coefficients are integers. In that scenario, there is a much simpler method.

$$ax^2+bx+c=(nx+p)(mx+q)$$

And you also have $nm=a$, $pq=c$, and $nq+mp=b$.

Just remember that the first constants on the right multiplied equals the first coefficient on the left. The last constants on the right multiplied equals the last constant on the left.

And the 'cross' multiplication equals the middle coefficient. 