The Product-Sum Technique

The product-sum technique makes it possible to factor a trinomial of the form $ax^2+bx+c.$

To factor a trinomial of the form $\color{green}{a}x^2 + \color{blue}{b}x + \color{green}{c}$ with the product-sum technique, follow these steps.

1. Find two numbers $\color{red}{m}$ and $\color{red}{n}$ whose product equals the value of $\color{green}{a}$ multiplied by $\color{green}{c}$, and whose sum equals the value of $\color{blue}{b}$.
   $$\color{green}{\text{ Product }} =\color{green}{a}\color{green}{c}\ \ \ \ \ \ \color{blue}{\text{Sum }}=\color{blue}{b}$$

2. Decompose the term $bx$ in the trinomial using the two numbers found.

3. Factor by grouping.

This method can be difficult if the values of $a$, $b$, and $c$ are fractions or fairly large integers (both positive and negative).

Moreover, if the discriminant of the trinomial $ax^2+bx+c$ is negative, it is not possible to factor the trinomial at all. $$b^2-4ac<0 \ \Longrightarrow \text{not factorable}$$

Consider the trinomial $x^2 + 4x – 32.$

1. Find the product and sum
   Identify the parameters $a$, $b$, and $c$ of this trinomial.

   $$\color{green}{a} = \color{green}{1},\ \color{blue}{b} = \color{blue}{4},\ \color{green}{c} = \color{green}{-32}$$ $$\begin{align}\color{green}{\text{Product}}&=\color{green}{a}\color{green}{c}& \color{blue}{\text{Sum}} &= \color{blue}{b}\\ &= \color{green}{1}\ \times\color{green}{-32}&&=\color{blue}{4} \\ &=\color{green}{-32}\end{align}$$

   We are looking for two numbers with a product of $-32$ and a sum of $4$. Use trial and error to determine them, like so:

   $$\begin{align}-1\times 32&=\color{green}{-32},\ \text{but}\ \ -1+32=31\\ 1\times -32&=\color{green}{-32}, \ \text{but}\ \ \ 1+(-32)=-31\\-2\times 16&=\color{green}{-32}, \ \text{but}\ \ -2+16=14\\ 2\times -16&=\color{green}{-32}, \ \text{but}\ \ \ 2+(-16)=-14\\ \color{red}{-4}\times \color{red}{8}&=\color{green}{-32}\ \ \ \text{and}\ \ \color{red}{-4}+\color{red}{8}=\color{blue}{4}\end{align}$$ The two numbers are therefore $\color{red}{-4}$ and $\color{red}{8}$.

2. Decompose the term $bx$ in the trinomial using the two numbers found

   $$\begin{align}x^2+4x-32&=x^2\color{red} {+4x}-32\\ &=x^2\color{red} {-4x+8x}-32\end{align}$$

3. Factor by grouping

   $$\begin{align}x^2+4x-32&=x^2-4x+8x-32\\&=x(x-4)+8(x-4)\\&=(x-4)(x+8)\end{align}$$

Consider the trinomial $6x^2+16x+8.$

1. Find the product and sum
   Before trying the product-sum technique, notice the polynomial has common factors. It is possible to factor out a common factor. $$6x^2+16x+8\\ 2(3x^2+8x+4)$$ Next, apply the product-sum technique to the trinomial $3x^2+8x+4$.
   
   Identify the parameters $a$, $b$, and $c$ of the trinomial. $$\color{green}{a} = \color{green}{3}, \color{blue}{b} = \color{blue}{8}, \color{green}{c} = \color{green}{4}$$ $$\begin{align}\color{green}{\text{Product}}&=\color{green}{a}\color{green}{c}& \color{blue}{\text{Sum}} &= \color{blue}{b}\\ &= \color{green}{3}\times \color{green}{4}&&=\color{blue}{8} \\ &=\color{green}{12}\end{align}$$ Look for two numbers with a product of $12$ and a sum of $8$.
   Use trial and error to determine them, as follows. $$\begin{align}1\times 12&=\color{green}{12},\ \text{but}\ \ 1+12=13\\ 3\times 4&=\color{green}{12}, \ \text{but}\ \ \ 3+4=7\\ \color{red}{2}\times \color{red}{6}&=\color{green}{12}\ \ \ \text{and}\ \ \color{red}{2}+\color{red}{6}=\color{blue}{8}\end{align}$$ The two numbers are therefore $\color{red}{2}$ and $\color{red}{6}$.

2. Decompose the term $bx$ in the trinomial using the two numbers found  $$\begin{align}6x^2+16x+8&=2(3x^2\color{red} {+8x}+4)\\ &=2(3x^2\color{red} {+2x+6x}+4)\end{align}$$

3. Factor by grouping  $$\begin{align}6x^2+16x+8&=2(3x^2+2x+6x+4)\\&=2\left(x(3x+2)+2(3x+2)\right)\\&=2(3x+2)(x+2)\end{align}$$