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Expanding Products of Linear Binomials

Good to see you again. Last time, you practiced evaluating a quadratic rule by substituting an input carefully, especially when the input is negative. That skill tells you what a quadratic does at particular $x$ -values. This lesson focuses on an earlier algebraic step that often produces quadratic rules in the first place: multiplying two binomials.

A binomial has two terms, such as $x + 2$ or $x - 5$ . By the end of this lesson, you will be able to expand products of the form

(x + p) (x + q)

into standard quadratic form,

x^{2} + b x + c .

This will matter later because factored and expanded forms reveal different useful features of the same parabola.

Parentheses mean multiplication

When two expressions are written next to each other in parentheses,

(x + p) (x + q),

they are being multiplied. The key rule is the distributive property: every term in the first binomial must multiply every term in the second binomial.

Begin by distributing the first $x$ :

x (x + q) = x^{2} + q x .

Then distribute $p$ :

p (x + q) = p x + pq .

Put those results together:

(x + p) (x + q) = x^{2} + q x + p x + pq = x^{2} + (p + q) x + pq .

So the general pattern is

(x + p) (x + q) = x^{2} + (p + q) x + pq .

There are always four individual products before simplification:

$x \cdot x$ , which gives $x^{2}$
$x \cdot q$ , which gives $q x$
$p \cdot x$ , which gives $p x$
$p \cdot q$ , which gives $pq$

The two middle products are both $x$ -terms, so they are like terms and can be combined. The first and last products have different variable parts, so they cannot be combined with the middle terms.

See the multiplication as area

An area model gives a visual reason that there must be four products. Imagine a rectangle with side lengths $x + 2$ and $x + 4$ . Split each side at the boundary between its variable and numerical parts. The large rectangle becomes four smaller rectangles.

An area model for $(x+2)(x+4)$: the four sub-rectangles have areas $x^2$, $2x$, $4x$, and $8$, which combine to give $x^2+6x+8$.

The total area is the sum of all four smaller areas:

(x + 2) (x + 4) = x^{2} + 4 x + 2 x + 8 = x^{2} + 6 x + 8.

This model explains a common mistake: writing only $x^{2} + 8$ misses the two rectangular regions with areas $4 x$ and $2 x$ . Those middle terms are not optional; they arise from genuine cross-products.

The same structure works even when a number is negative. For example:

(x - 3) (x + 5) = (x + (- 3)) (x + 5) = x^{2} + 5 x - 3 x - 15 = x^{2} + 2 x - 15.

Treat $x - 3$ as $x + (- 3)$ . This makes the signs much easier to track: $(- 3) (5) = - 15$ .

FOIL: a useful bookkeeping shortcut

For two binomials only, the distributive process is often remembered with FOIL:

First terms
Outer terms
Inner terms
Last terms

FOIL is not a separate mathematical rule. It is simply a way to make sure all four required products are included.

Example 1: Multiplying a binomial by a binomial | Algebra I | Khan Academy

Watch “Example 1: Multiplying a binomial by a binomial” from Khan Academy. It first demonstrates FOIL, then shows why distributing every term gives the same result.

Watch the FOIL setup to see how the four term-pairs are identified in (3x+2)(5x-7). Then watch the simplification, focusing on why only the two x-terms combine. Finish with the distributive explanation, which is the more reliable method when expressions become less familiar.

Consider

(x - 6) (x - 2) .

Using FOIL, write each product with its sign:

Pair	Product
First: $x \cdot x$	$x^{2}$
Outer: $x \cdot (- 2)$	$- 2 x$
Inner: $(- 6) \cdot x$	$- 6 x$
Last: $(- 6) \cdot (- 2)$	$12$

Now add the four products and combine like terms:

(x - 6) (x - 2) = x^{2} - 2 x - 6 x + 12 = x^{2} - 8 x + 12.

Notice the role of signs:

A positive times a negative is negative.
A negative times a positive is negative.
A negative times a negative is positive.

It is safer to view the terms as $x$ , $- 6$ , $x$ , and $- 2$ , rather than thinking of subtraction marks as instructions to perform later.

A fast pattern for the specific form $(x + p) (x + q)$

Because both leading terms are $x$ , the expansion has a predictable structure:

(x + p) (x + q) = x^{2} + (p + q) x + pq .

This means:

the coefficient of $x^{2}$ is always $1$ ;
the coefficient of $x$ is the sum of the two constants;
the final constant is the product of the two constants.

For example,

(x + 7) (x - 3)

has constants $7$ and $- 3$ . Their sum is $4$ , and their product is $- 21$ . Therefore,

(x + 7) (x - 3) = x^{2} + 4 x - 21.

You can still write all four products if you prefer:

(x + 7) (x - 3) = x^{2} - 3 x + 7 x - 21 = x^{2} + 4 x - 21.

The shortcut is useful, but the full distribution is the method to return to whenever signs feel uncertain.

Check an expansion by testing one value

An expanded expression and its original product should give the same output for every $x$ . Evaluating both at one convenient value is a quick way to detect many slips.

Suppose you expanded

(x - 4) (x - 7)

x^{2} - 11 x + 28.

Test $x = 1$ .

Using the original product:

(1 - 4) (1 - 7) = (- 3) (- 6) = 18.

Using the expansion:

1^{2} - 11 (1) + 28 = 1 - 11 + 28 = 18.

Both forms give $18$ , so this check supports the expansion.

This does not replace careful multiplication, but it connects directly to the function-evaluation work from the previous lesson: equivalent algebraic forms must produce identical outputs for the same input.

Common errors to avoid

Forgetting the middle terms

(x + 2) (x + 4) \neq = x^{2} + 8.

The correct expansion includes both cross-products:

(x + 2) (x + 4) = x^{2} + 4 x + 2 x + 8 = x^{2} + 6 x + 8.

Combining unlike terms

x^{2} + 5 x - 3 x - 15,

only $5 x$ and $- 3 x$ combine. You cannot combine $x^{2}$ with $x$ , and you cannot combine either variable term with $- 15$ .

x^{2} + 5 x - 3 x - 15 = x^{2} + 2 x - 15.

Losing the variable in the middle term

The sum $p + q$ is a number, but it is the coefficient of $x$ :

(x + p) (x + q) = x^{2} + (p + q) x + pq,

not

x^{2} + (p + q) + pq .

Treating a square incorrectly

When the two binomials happen to match, distribute as usual:

(x + 3)^{2} = (x + 3) (x + 3) = x^{2} + 3 x + 3 x + 9 = x^{2} + 6 x + 9.

In particular,

(x + 3)^{2} \neq = x^{2} + 9.

The middle term comes from the two cross-products.

Key takeaways

To expand two linear binomials, multiply each term in one binomial by each term in the other, then combine like terms. For the form in this lesson,

(x + p) (x + q) = x^{2} + (p + q) x + pq .

FOIL can help organize the four products, but the distributive property is the reason the process works. Keep signs attached to their terms, write all four products before combining, and use a quick input-value check if you want to verify the result.

Next, you will reverse this process: given a quadratic such as