Expanding and Factoring Expressions

Learning goals

Expand a product so the factor reaches every term
Multiply two expressions by pairing every term with every term
Treat FOIL as a name for four products, not a general rule
Factor by pulling out the greatest common factor
Check any factoring by re-expanding it

Expanding a factor across every term

To expand a product is to multiply it out into a sum, using the distributive law. You met the law for a sum of two terms, $a(b + c) = ab + ac$ , but nothing about it stops at two terms. The factor out front reaches every term inside the parentheses, however many there are.

Why $a(b + c + d) = ab + ac + ad$ #

The distributive law is stated for a sum of two terms, $a(b + c) = ab + ac$ , but a sum can have more terms than that. To see that the factor still reaches each one, group the extra terms and apply the two-term law twice. Treat $b + c + d$ as $(b + c) + d$ , a sum of the single quantity $b + c$ and the term $d$ :

a\big((b + c) + d\big) = a(b + c) + ad.

That is the two-term law with $b + c$ playing the role of the first term. Now distribute across the remaining pair:

a(b + c) + ad = ab + ac + ad.

The factor $a$ has multiplied every term inside. The same grouping works for a sum of any length, so a factor outside a parenthesis multiplies each term within it, one at a time. That is the whole content of expanding: no term is ever skipped.

The factor out front need not be a bare number. It can carry a sign or a variable, and the rule does not change: multiply it against each term, following the sign and exponent rules you already know. When two powers of the same base meet, their exponents add, so $x \cdot x = x^2$ and $x \cdot x^2 = x^3$ . A negative factor flips the sign of each term it lands on, exactly as a leading minus did in the last lesson. The flip happens because a minus sign in front is the factor $-1$ waiting to be distributed.

Worked example 1 Expand three products

In each product, multiply the outside factor against every inside term, keeping each sign.

Distributing a negative number across a difference, the sign of each term flips:

-3(x - 4) = (-3)(x) + (-3)(-4) = -3x + 12.

Distributing a single variable, the powers add where the variable meets itself:

x(2x + 3) = x \cdot 2x + x \cdot 3 = 2x^2 + 3x.

Distributing a term that carries both a coefficient and a power, handle the numbers and the exponents separately:

2x^2(3x - 5) = 2x^2 \cdot 3x - 2x^2 \cdot 5 = 6x^3 - 10x^2.

In each case the factor reached both terms. The second term is never left behind.

One case deserves a second look, because the factor is hidden. Subtracting a whole group, as in $a - (b + c)$ , is the same as multiplying that group by $-1$ and adding, so every term inside changes sign:

a - (b + c) = a - b - c, \qquad -(a - b) = -a + b.

This is the leading-minus rule from the last lesson seen as one more case of expanding, with the hidden factor being $-1$ .

Check your understanding

Expand $-2x(3x - 4)$ .

-6x^2 - 8x

-6x + 8

-6x^2 + 8x

6x^2 - 8x

Multiplying two expressions

Multiplying two multi-term expressions, such as $(a + b)(c + d)$ , looks like a new problem, but a picture makes it concrete first. Give a rectangle a width of $a + b$ and a height of $c + d$ . Splitting the width into $a$ and $b$ , and the height into $c$ and $d$ , carves the rectangle into four smaller rectangles, and their areas are exactly the four products you get from multiplying every part of one sum by every part of the other.

A rectangle with width a plus b and height c plus d splits into four smaller rectangles with areas ac, bc, ad, and bd. Their total is the whole area, so the product of the two sums equals the sum of the four pairwise products.

The same result follows from the distributive law alone, with no picture needed. Treat the first factor as a single quantity for a moment: nothing in the distributive law requires the outside factor to be short, so let the whole of $a + b$ play that role.

Why $(a + b)(c + d) = ac + bc + ad + bd$ #

Nothing in the distributive law says the outside factor has to be a single letter. Treat the whole quantity $a + b$ as one factor and distribute it across $c + d$ , the usual way:

(a + b)(c + d) = (a + b)c + (a + b)d.

Each of those two pieces is itself a distributive-law step, this time with $c$ and $d$ spreading back across $a + b$ :

(a + b)c = ac + bc, \qquad (a + b)d = ad + bd.

Adding the four pieces gives

(a + b)(c + d) = ac + bc + ad + bd.

Every term of the first factor has multiplied every term of the second, one product for each pairing, matching the four cells of the rectangle above. Because the argument used nothing but the distributive law, it does not stop at two terms each. So to multiply any two expressions, multiply every term of the first by every term of the second, then combine like terms.

For the common case of two two-term factors, the four products come in a memorable order. Reading $(a + b)(c + d)$ left to right, they are the product of the First terms $ac$ and the Outer terms $ad$ . Then come the Inner terms $bc$ and the Last terms $bd$ , so the four initials spell the mnemonic FOIL. It is a handy way to be sure you have all four products, but it is only a name for what the distributive law already told you. FOIL works only when each factor has exactly two terms. The underlying rule, every term times every term, is what you fall back on the moment a factor has three terms or more.

Worked example 2 Multiply two binomials

Multiply every term of the first factor by every term of the second, then gather like terms.

For $(x + 2)(x + 3)$ , the four products are $x \cdot x$ , $x \cdot 3$ , $2 \cdot x$ , and $2 \cdot 3$ :

(x + 2)(x + 3) = x^2 + 3x + 2x + 6.

The middle terms $3x$ and $2x$ are like terms, so combine them:

x^2 + 3x + 2x + 6 = x^2 + 5x + 6.

The same method handles coefficients and negatives. For $(2x - 1)(x + 4)$ ,

(2x - 1)(x + 4) = 2x^2 + 8x - x - 4 = 2x^2 + 7x - 4,

where $2x \cdot x = 2x^2$ , the outer and inner products $8x$ and $-x$ combine to $7x$ , and $(-1)(4) = -4$ .

Check your understanding

Multiply $(x + 4)(x + 2)$ .

x^2 + 8

x^2 + 6x + 8

x^2 + 8x + 8

x^2 + 6x + 6

Worked example 3 Multiply a binomial by a trinomial

When a factor has three terms, FOIL no longer applies, but every term times every term still does. Multiply each of the two terms of $x + 2$ against all three terms of $x^2 + 3x - 1$ :

(x + 2)(x^2 + 3x - 1) = x(x^2 + 3x - 1) + 2(x^2 + 3x - 1).

Expand each product:

x(x^2 + 3x - 1) = x^3 + 3x^2 - x, \qquad 2(x^2 + 3x - 1) = 2x^2 + 6x - 2.

Add them and combine the like $x^2$ terms and the like $x$ terms:

x^3 + 3x^2 - x + 2x^2 + 6x - 2 = x^3 + 5x^2 + 5x - 2.

That is six products in all, one for each pairing of a term from the first factor with a term from the second.

Check your understanding

How many products does multiplying $(x^2 + 4x - 3)(x + 5)$ require, before combining any like terms?

A Five, one for each term in the two factors combined B Six, one for each pairing of a term from one factor with a term from the other C Three, one for each term in the trinomial D Four, using FOIL

Factoring by reversing distribution

Expanding turns a product into a sum. Factoring runs the machine backward. In this lesson, that means turning a sum into a product by pulling out a factor shared by every term. The tool is the distributive law read from right to left, $ab + ac = a(b + c)$ , which says a factor common to each term can be lifted outside a parenthesis.

Why pulling out a common factor is the distributive law reversed#

The distributive law, read left to right, expands: $a(b + c) = ab + ac$ . Every equation can be read in either direction, so read this one right to left:

ab + ac = a(b + c).

This says that if a factor $a$ appears in every term of a sum, it can be pulled out to the front. The other factors of each term stay behind inside the parenthesis. Nothing new is needed; it is the same identity, used in reverse.

To factor $6x + 9$ , look for a factor common to both terms. Since $6x = 3 \cdot 2x$ and $9 = 3 \cdot 3$ , the number $3$ divides each, so lift it out:

6x + 9 = 3 \cdot 2x + 3 \cdot 3 = 3(2x + 3).

For $x^2 + 3x$ the shared factor is the variable $x$ , because $x^2 = x \cdot x$ and $3x = x \cdot 3$ :

x^2 + 3x = x \cdot x + x \cdot 3 = x(x + 3).

In both cases you can check the result at once by expanding it again: $3(2x + 3) = 6x + 9$ and $x(x + 3) = x^2 + 3x$ , right back to where you started.

The factor worth pulling out is the greatest common factor, or GCF: the largest factor every term shares. It has two parts. The numerical part is the greatest common divisor of the coefficients, and the variable part is each common variable raised to the smallest power that appears. For $4x^2 + 6x$ the coefficients $4$ and $6$ share the factor $2$ , and both terms contain at least one $x$ , so the GCF is $2x$ . Pulling it out,

4x^2 + 6x = 2x \cdot 2x + 2x \cdot 3 = 2x(2x + 3).

Taking out less than the full GCF leaves the job unfinished. The forms $2(2x^2 + 3x)$ and $x(4x + 6)$ are both correct as far as they go, but each still hides a shared factor inside, so neither is fully factored.

The area model reads backward here too. A rectangle whose area is $6x + 9$ and whose height is the common factor $3$ must have width $2x + 3$ . So factoring is just naming the side lengths of the rectangle you already drew when expanding.

A rectangle of height 3 and width 2x plus 3 has area 6x plus 9, split into a part of area 6x and a part of area 9. Reading the common height 3 as a factor pulled out front shows 6x plus 9 equals 3 times the quantity 2x plus 3.

Worked example 5 Factor out the greatest common factor

Find the GCF, lift it out, then confirm the answer by expanding it again.

For $12x^2 - 8x$ , the coefficients $12$ and $8$ share $4$ and both terms contain $x$ , so the GCF is $4x$ :

12x^2 - 8x = 4x(3x - 2).

Expanding the answer returns the original, which is the check that the factoring is right: $4x(3x - 2) = 12x^2 - 8x$ , as it should.

Check your understanding

Factor out the greatest common factor of $8x^2 + 12x$ .

2x(4x + 6)

4(2x^2 + 3x)

4x(2x + 3)

x(8x + 12)

Check your understanding

To check a proposed factoring, you expand it back out and compare. Expand $3x(3x + 2)$ using the distributive law. What do you get?

9x^2 + 6x

9x^2 + 2x

3x^2 + 2x

6x^2 + 6x

Common mistakes

Distributing to only the first term. A factor must reach every term inside, so $x(x + 5) = x^2 + 5x$ , not $x^2 + 5$ , and $3(2x - 4) = 6x - 12$ , not $6x - 4$ .
Dropping the middle terms of a product. Multiplying two binomials makes four products, and the two middle ones combine only when they are like terms, as they are here: $(x + 2)(x + 3) = x^2 + 5x + 6$ , not $x^2 + 6$ .
Mishandling a sign on a negative factor. A negative factor flips the sign of every term: $-2(x - 3) = -2x + 6$ , not $-2x - 6$ , and $-(x - 4) = -x + 4$ .
Squaring a sum term by term. $(x + 3)^2$ is $(x + 3)(x + 3) = x^2 + 6x + 9$ , not $x^2 + 9$ . The middle term $6x$ comes from the two cross products and is the piece most often forgotten.
Pulling out only part of the GCF. $8x^2 + 12x = 4x(2x + 3)$ . Writing $2x(4x + 6)$ leaves a shared factor of $2$ still inside, so the expression is not yet fully factored.
Confusing $x \cdot x$ with $x + x$ . Multiplying gives $x \cdot x = x^2$ , while adding gives $x + x = 2x$ . So $x(x + 1) = x^2 + x$ , where the first term is a power and the second is not.

Key takeaways

To expand is to multiply out a product with the distributive law, so a factor reaches every term: $a(b + c + d) = ab + ac + ad$ , and $-3(x - 4) = -3x + 12$ .
To multiply two expressions, multiply every term of the first by every term of the second, then combine like terms: $(a + b)(c + d) = ac + bc + ad + bd$ , so $(x + 2)(x + 3) = x^2 + 5x + 6$ .
FOIL names the four products (First, Outer, Inner, Last) for two binomials only; the rule every term times every term is what handles three terms or more.
To factor is to run distribution in reverse, pulling out the greatest common factor. That factor is the greatest common divisor of the coefficients times each common variable at its lowest power. So $6x + 9 = 3(2x + 3)$ and $4x^2 + 6x = 2x(2x + 3)$ .
Check any factoring by re-expanding. If the product does not return the original expression, a factor or a sign is off.

Expanding and Factoring Expressions

Learning goals

Expanding a factor across every term

Multiplying two expressions

Factoring by reversing distribution

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice