FOIL and quadratic equations

FOILing

FOILing is a tactic you can use to rewrite quadratics. Sometimes a problem will have two parts in parentheses multiplied by each other:

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

You can simplify this expression with the FOIL method:

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

To start, the first part of the left parenthesis must be multiplied by the first part of the right parenthesis. In our example, this is $(x)(x)=x^2$. Next, we’ll multiply the outside parts: $(x)(-2)=-2x$. Then, the inside and last parts. Once we have all the solutions, add them together to get the simplified version of the quadratic.

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 equation $(x+3)(x-2)$ could be represented as $x^2+x-6$.

Reverse FOILing (factoring quadratic equations)

FOILing is done to expand binomial terms into a full-blown quadratic equation, and reverse FOILing can be done if we need to go in the opposite direction. For example, $x^2+5x-6=0$ can be reverse FOILed (a.k.a. factored) into $(x+6)(x-1)=0$.

This technique is most commonly used when the quadratic is equal to 0. If two parenthetic binomials equal 0 when multiplied together, one or both of them must equal zero. By setting both equal to 0, you can determine the two possible values of the variable.

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

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

Now that we know the purpose of reverse FOILing, let’s walk through the simplest way to do it. We’ll use the same example quadratic equation as before:

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

To reverse FOIL a polynomial that does not have a coefficient in front of the squared variable, always start by asking yourself what two numbers have a product of the last part of the sequence, i.e. the part is just a simple constant number without an $x$. The pair of numbers that have a product of $-6$ are $(-1,6)$, $(6,-1)$, $(3,-2)$, and $(-3,2)$. Which of these pairs sum up to the middle number, $5$? Since $-1+6=5$, we can select this pair of numbers and insert $-1$ and $6$ into the two binomials.

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

Pretty straightforward, right? Just start by finding the pairs that have a product of the last number, then determine which pair has a sum of the middle number.

If there is a coefficient in front of the first part of the polynomial, it becomes a little more complicated, but we can use the same core strategy. Let’s try factoring this more complex equation:

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

The final non-$x$ term is $-15$, so we’ll start by listing out all the pairs of integers that have a product of $-15$: $(5,-3)$, $(-5,3)$, $(15,-1)$, and $(-15,1)$.

Because our polynomial starts with $2x^2$ and the middle term is $-7x$, we need to ask ourselves which pair of numbers will have a sum of $-7$ if one of the numbers in the pair is multiplied by $2$.

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

Now that we’ve figured out the terms, we just need to be careful to put them in the right place. When we FOIL we want it to end up that we’re multiplying $2\ast -5$, so that means that we need to have our $2x$ First term separate from our $-5$ Last term:

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

If it’s tricky to think through, you can always just try it both ways. There are only two possible ways to arrange it, so even using brute force you’ll figure out the right order quickly.

Sidenote

Factoring BIG quadratic equations

Sometimes a problem requires you to reverse FOIL an equation that has numbers that would seem too large to use the strategies described above. If each number in the polynomial shares a common multiple, try dividing the whole equation by that number.

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: $x^2+2x-3=0$.

Dividing out a constant number from the entire equation before reverse FOILing works because the right side of the equation stays the same. 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 you’re stuck, you can use the quadratic formula. This tends to be less relevant for the GRE, but it’s good to know just in case.

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

Let’s use this formula to solve the previous quadratic equation, $2x^2-7x-15=0$.

The first step is to match the variables to their values in the equation. Here’s the template for it:

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

Our equation is $2x^2-7x-15=0$, which means we get the following values:

• $a=2$
• $b=-7$
• $c=-15$

Now we can plug this into the quadratic equation and solve for $x$. Remember how quadratic equations have two roots? This corresponds with the plus/minus $\pm$ sign in the formula. It means that you need to solve twice, once adding the following part, and once subtracting the following part, and both answers are roots to the equation.

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

We’ll change the plus/minus $\pm$ to a plus $+$ to solve for the first root:

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

And we’ll change the plus/minus $\pm$ to a minus $-$ to solve for the second root:

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

Note that this is the same solution you would find with the reverse FOIL method!

In summary, the quadratic formula isn’t commonly used in the GRE. However, it might be relevant if the quadratic involves larger/complex numbers or you can’t easily reverse FOIL.

Bringing it all together: question walkthrough video

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