FOIL and quadratic equations

FOILing

FOILing is a method you can use to expand (multiply out) two binomials. Sometimes you’ll see an expression with two parenthetical terms multiplied together, like this:

$$(x+3)(x-2)$$

You can simplify this expression using the FOIL method:

1. First
2. Outside
3. Inside
4. Last

Here’s what each step means. You multiply specific pairs of terms from the two parentheses:

• First: first term in the left parenthesis × first term in the right parenthesis
• Outside: outside terms
• Inside: inside terms
• Last: last term in the left parenthesis × last term in the right parenthesis

After you multiply each pair, you add the results and combine like terms.

Here’s the full FOIL process step by step:

1. First

$$(x)(x)=x^2$$

2. Outside

$$(x)(-2)=-2x$$

3. Inside

$$(3)(x)=3x$$

4. Last

$$(3)(-2)=-6$$

5. Add everything together to get the result

$$x^2-2x+3x-6=x^2+x-6$$

So, the expression $(x+3)(x-2)$ can be rewritten as $x^2+x-6$.

Reverse FOILing (factoring quadratic equations)

FOILing expands binomials into a quadratic. Reverse FOILing goes the other direction: it rewrites a quadratic as a product of two binomials (this is also called factoring).

For example, you can factor

$$x^2+5x-6=0$$

into

$$(x+6)(x-1)=0.$$

This technique is most commonly used when the quadratic is set equal to 0. If two binomials multiply to 0, then at least one of them must be 0. So you set each factor equal to 0 to find the possible values of $x$.

$$(x+6)=0\text{OR}(x-1)=0$$

These solutions are also called the “roots of the equation”: $x$ is either $-6$ or $1$.

Now that you know what reverse FOILing is for, here’s a straightforward way to do it. We’ll use the same example:

$$x^2+5x-6=0$$

When the coefficient of $x^2$ is 1 (so the quadratic starts with $x^2$), start by looking at the constant term (the last term, with no $x$). Here the constant term is $-6$.

1. List pairs of integers whose product is $-6$:

$(-1,6)$, $(6,-1)$, $(3,-2)$, and $(-3,2)$.

2. Choose the pair that adds to the middle coefficient, $5$.

Since $-1+6=5$, use $-1$ and $6$.

3. Write the factors:

$$(x+6)(x-1)=0$$

So the process is:

• Find pairs that multiply to the constant term.
• Pick the pair that adds to the $x$ coefficient.

If there is a coefficient in front of the squared term, it becomes a little more complicated, but the core idea is similar. Let’s factor:

$$2x^2-7x-15=0$$

The constant term is $-15$, so start by listing pairs of integers whose product is $-15$:

$(5,-3)$, $(-5,3)$, $(15,-1)$, and $(-15,1)$.

Because the quadratic starts with $2x^2$ and the middle term is $-7x$, you want a pair where one number gets multiplied by 2 (from the $2x^2$) and then the two results add to $-7$.

The correct choice is $(-5,3)$, since $2(-5)+3=-7$.

Now place the terms so that FOILing produces the correct middle term. You want the First × Last product to be $2x\cdot (-5)$, so the $2x$ and the $-5$ must be in different parentheses:

$$(2x+3)(x-5)=2x^2-7x-15=0$$

If the placement feels hard to see, you can check by FOILing. There are only two possible arrangements, so testing both is quick.

Sidenote

Factoring BIG quadratic equations

Sometimes you need to reverse FOIL an equation with numbers that feel too large to handle directly. If every term shares a common factor, factor it out first.

For example:

$$\begin{array}{rl}4x^2+8x-12& =0\\ 4\ast (x^2+2x-3& =0)\end{array}$$

And now this is manageable to factor: $x^2+2x-3=0$.

Factoring out a constant works here because the right side stays 0. After all, 0 divided by anything is still 0.

The quadratic formula

Reverse FOILing is usually the fastest way to solve a quadratic equation, but if factoring isn’t working, you can use the quadratic formula. This tends to be less relevant for the GRE, but it’s good to know.

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Let’s use this formula to solve the previous quadratic equation:

$$2x^2-7x-15=0.$$

First, match the equation to the standard form:

$$a{x}^2+bx+c=0$$

Our equation is $2x^2-7x-15=0$, so:

• $a=2$
• $b=-7$
• $c=-15$

Now plug these values into the quadratic formula. The $\pm$ means you’ll get two solutions (two roots): one using $+$ and one using $-$.

$$\begin{array}{rl}x& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x& =\frac{-(-7)\pm \sqrt{(-7{)}^2-4(2)(-15)}}{2(2)}\\ x& =\frac{7\pm \sqrt{49+120}}{4}\\ x& =\frac{7\pm \sqrt{169}}{4}\\ x& =\frac{7\pm 13}{4}\end{array}$$

Use $+$ to find the first root:

$$\begin{array}{rl}x& =\frac{7+13}{4}\\ x& =\frac{20}{4}\\ x& =5\end{array}$$

Use $-$ to find the second root:

$$\begin{array}{rl}x& =\frac{7-13}{4}\\ x& =\frac{-6}{4}\\ x& =\frac{-3}{2}\end{array}$$

Notice that these are the same solutions you would get by reverse FOILing.

In summary, the quadratic formula isn’t commonly used on the GRE, but it’s useful when the quadratic doesn’t factor easily or involves larger/complex numbers.

Bringing it all together: question walkthrough video

Here’s a video walking through one of our practice questions to show these ideas in action: