Factoring Quadratics where $a=1$

Precalculus Algebra Factoring Quadratic Expressions Factoring Quadratics where $a=1$

The Method

Recall that a quadratic expression is a degree $2$ polynomial of the form

$$\begin{align} ax^2+bx+c\end{align}$$

where the coefficients $a,b,c$ are all real numbers.

Here, we focus our attention on factoring quadratic expressions of the form

$$\begin{align}x^2+bx+c\end{align}$$

that is, where the leading coefficient is $a=1$.

How to Factor Quadratics where $a=1$.

Starting with an expression of the form $x^2+bx+c$

1. Find two numbers that add together to get $b$ and also multiply together to get $c$. Call them $n$ and $m$.

2. Use $m$ and $n$ to “break up the middle term” so that
$$\begin{align} x^2+bx+c=x^2+nx+mx+c\end{align}$$

3. Factor the expression by grouping the first pair of terms and the last pair.

4. After factoring by grouping, you should end up with an expression of the form
$$\begin{align}(x + n)(x +m)\end{align}$$

This method “undoes” the process of distribution. To see this, take the result above and FOIL it. As you simplify, you will see that you are working the steps of the method above in reverse.

Following the process above, we need to find two numbers $n$ and $m$ that when added together give you $b=2$ and when multiplied together give you $c=1$.

Such numbers are $n=2$ and $m=3$ (since $2\cdot 3 =6=c$ and $2+3=5=b$).

Now break apart the middle term using these numbers and then factor by grouping.

$$\begin{align}x^2+5x+6&=x^2+2x+3x+6\\&= (x^2+2x)+(3x+6)\\&=x(x+2)+3(x+2)\\&=(x+2)(x+3)\end{align}$$

Why the Method Works

Take a look at the following distribution/expanding problem, and try to work it out yourself real quick.

Solution:

$$\begin{align} (x+2)(x+3)=x^2+3x+2x+6=x^2+5x+6\end{align}$$

Notice that in the first step of the above problem we have the pair of middle terms (namely $3x$ and $2x$) which came from multiplying the $x$’s in the first and second quantity to the numbers $2$ and $3$ as part of the distribution process. Adding these together gives you the middle term in the final answer. Multiplying those same two numbers $2$ and $3$ gives you $6$ as well (as part of the distribution process).

So, as part of the distribution process, the $2$ and the $3$ end up being added together to give you the coefficient $b=5$ on the $x$, and $2$ and $3$ multiplied together give you $c=6$.

Now, if instead we were asked to reverse the process of distribution, starting with $x^2+5x+6$ with the goal of getting to a form like $(\ \ \ +\ \ \ )(\ \ \ +\ \ \ )$, we try to find two numbers that add up to $b=5$ and that when multiplied together give you $c=6$ as a way of “undoing” the distribution that you see above. Once we have such numbers (namely $2$ and $3$), we can split up the middle term of $x^2+5x+6$, thus allowing us to reverse the process of distribution above, by factoring by grouping:

$$\begin{align}x^2+5x+6&=x^2+2x+3x+6\\&=(x^2+2x)+(3x+6)\\&=x(x+2)+3(x+2)\\&=(x+2)(x+3)\end{align}$$

So, how one arrives at the values for $b$ and $c$ as part of the distribution process tells you how to find the two numbers that add to get $b$ and multiply to get $c$ as part of the factoring process.

A Few Caveats

WARNING: The above method only works when you have a $1$ in front of your $x^2$. Otherwise, you will need to use an altered version of this method (see next topic).

ANOTHER WARNING: Not every quadratic expression can be factored using this method, or the one that follows in the next topic. ALL Quadratics technically can be factored using the quadratic formula (something we will talk about in future lessons).