Multiplying Polynomials

Intuition

You can already add and subtract polynomials by collecting like terms. Multiplying them is the operation behind almost everything still ahead. That work includes expanding a product before you solve an equation, rewriting a function so you can graph it, and computing an area whose sides are themselves expressions. The good news is that there is no new machinery to learn. Every product of polynomials, however long, is the distributive property applied as many times as it takes, “each term times each term.” That expansion is followed by the same collecting of like terms you just practiced. The special products you met while factoring, such as $(a + b)^2$ and $(a + b)(a - b)$ , are only this one rule with the like terms already gathered. FOIL, which you may have heard of, is just a nickname for the smallest case.

Learning goals

Multiply coefficients and add exponents for two terms
Distribute a single term across every term
Pair every term of one factor with every term of the other
Treat FOIL as the two-binomial case, not the rule
Add the degrees to get the degree of a product

Multiplying one term by one term

Before multiplying whole polynomials, start with the atom of the whole subject, a single term times a single term. A term is a coefficient times a power of $x$ , so a product of two terms is a product of two numbers and two powers. Multiplication lets you rearrange those factors into any order. Group the numbers together and the powers together:

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

The coefficients multiply the ordinary way, $3 \cdot 4 = 12$ . The powers combine by the product rule for exponents, which says $x^m \cdot x^n = x^{m+n}$ , so you add the exponents:

(3x^2)(4x^3) = 12\,x^{2+3} = 12x^5.

Multiply the coefficients, add the exponents. You add rather than multiply the exponents because $x^2 \cdot x^3$ is two $x$ ‘s times three more $x$ ‘s. That is five $x$ ‘s multiplied in a row, not six. When a term carries a negative sign, fold the sign into the coefficient and proceed exactly the same way:

(-5x^4)(2x) = (-5 \cdot 2)(x^4 \cdot x^1) = -10x^5.

That is the one small skill every product in this lesson is built from. Everything else is doing it many times and adding up the pieces.

Multiplying a polynomial by a single term

When one factor has several terms, the distributive property takes over. It says that a single factor multiplies each term of a sum separately:

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

The lone factor $a$ reaches every term inside the parentheses, not just the first. When $a$ and each of $b, c, d$ are themselves terms in $x$ , every one of those small products is a one-term-times-one-term computation you just practiced.

Worked example 1 Multiply $2x^3(4x^2 - 3x + 5)$

Distribute $2x^3$ across all three terms, keeping each sign with its term:

2x^3(4x^2 - 3x + 5) = (2x^3)(4x^2) + (2x^3)(-3x) + (2x^3)(5).

Now do each piece by multiplying coefficients and adding exponents:

(2x^3)(4x^2) = 8x^5, \qquad (2x^3)(-3x) = -6x^4, \qquad (2x^3)(5) = 10x^3.

Nothing here is a like term of anything else, so the product is already in standard form:

2x^3(4x^2 - 3x + 5) = 8x^5 - 6x^4 + 10x^3.

Multiplying two binomials

Now both factors are sums. The distributive property still does all the work; you simply use it twice. Hold the second binomial $(c + d)$ together as a single block for a moment. Then $(a + b)$ distributes across it, and each of the two pieces distributes again.

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

Treat the second factor $c + d$ as one quantity and distribute the sum $a + b$ over it, exactly as the distributive property allows:

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

The product has split into two monomial-times-binomial products, and each of those distributes in turn. The factor $a$ reaches both terms of $c + d$ , and so does the factor $b$ :

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

Putting the pieces back together gives four products,

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

and there is a plain pattern in them. Each of the two terms in the first factor has been multiplied by each of the two terms in the second. No term is special, and none is skipped. Every term meets every term, which is the same idea that will run the general rule below.

Those four products have a well-known nickname. Reading them in the order First, Outer, Inner, Last spells FOIL. The letters name, in order, the product of the First terms ( $ac$ ), the Outer terms ( $ad$ ), the Inner terms ( $bc$ ), and the Last terms ( $bd$ ). FOIL is a fine way to remember the four products of a binomial times a binomial, but it is not a separate rule. Its limit is also worth being clear about. FOIL names exactly the two-terms-times-two-terms case and nothing more. It says nothing about a binomial times a trinomial, where there are six products rather than four. The idea that does keep working, no matter how many terms each factor has, is the plain one: each term times each term. That is the rule to hold onto, with FOIL as a handy label for the special case where it produces four pieces.

A picture makes the four products concrete. If $x$ is a positive length, then $x + 2$ and $x + 3$ are side lengths, and the product $(x + 2)(x + 3)$ is the area of a rectangle with those sides. Splitting each side at its two parts cuts the rectangle into four smaller rectangles, and their four areas are exactly the four products.

An area model for the product (x + 2)(x + 3). The whole rectangle is x + 3 wide and x + 2 tall, so its area is the product. Splitting each side gives four smaller rectangles whose areas are the four partial products, and together they add to x squared plus 5x plus 6.

The two cross rectangles, $3x$ on the top right and $2x$ on the bottom left, are the Outer and Inner products. They carry the same power of $x$ , so they combine into the middle term, and this is the pair that is so easy to forget.

Worked example 2 Expand $(x + 2)(x + 3)$

Multiply each term of the first binomial by each term of the second, four products in all:

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

The two middle products are like terms, so combine them:

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

These are the four areas from the figure, $x^2$ , $3x$ , $2x$ , and $6$ , added together. Leaving the answer as $x^2 + 3x + 2x + 6$ is not wrong, only unfinished; combining like terms is what puts it in standard form.

Negative terms bring nothing new. Keep each sign attached to its term, and let the sign ride through every product.

Worked example 3 Expand $(2x - 3)(x + 4)$

The four products, with signs kept in place, are

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

Combine the two middle terms, $8x - 3x = 5x$ :

(2x - 3)(x + 4) = 2x^2 + 5x - 12.

The term $(-3)(x)$ is $-3x$ and $(-3)(4)$ is $-12$ . Dropping either minus sign would corrupt a whole term, which is the single most common slip in this kind of product.

Check your understanding

Expand $(x - 4)(x + 6)$ .

x^2 + 2x - 24

x^2 - 2x - 24

x^2 + 10x - 24

x^2 - 24

The general rule: every term times every term

Nothing about the method depended on each factor having exactly two terms. The same distribution works for any two polynomials, and it reads as one sentence:

To multiply two polynomials, multiply every term of the first by every term of the second, then combine like terms and write the result in standard form.

A factor with $m$ terms times a factor with $n$ terms produces $m \times n$ products before you combine anything. A binomial times a binomial gives $2 \times 2 = 4$ products, which is the case FOIL covers; a binomial times a trinomial gives $2 \times 3 = 6$ .

You have already seen this general rule specialized. Squaring a binomial is $(a + b)(a + b)$ , whose four products collapse to $a^2 + 2ab + b^2$ . The difference of squares $(a + b)(a - b)$ is the same four products with the two cross terms cancelling to leave $a^2 - b^2$ . Those familiar patterns from the special-factorizations chapter are not separate facts to memorize here; they are this one method with the like terms already collected. In this lesson you keep the method general.

With more than four products to track, a grid keeps every one of them in its own cell so none can be lost. Put the terms of one factor along the top and the terms of the other down the side. Then fill each cell with the product of its row and column headers.

The product (x + 3)(x^2 + 2x + 5) laid out as a grid, one row per term of x + 3 and one column per term of x squared + 2x + 5. Each of the six cells holds a single partial product, so none can be dropped. Adding the cells and combining like terms gives x cubed plus 5x squared plus 11x plus 15.

Worked example 4 Expand $(x + 3)(x^2 + 2x + 5)$

The binomial has two terms and the trinomial has three, so expect $2 \times 3 = 6$ products. Distribute $x$ across the trinomial, then $3$ across it:

x(x^2 + 2x + 5) = x^3 + 2x^2 + 5x, \qquad 3(x^2 + 2x + 5) = 3x^2 + 6x + 15.

These are the six cells of the grid. Add them and group like powers:

x^3 + 2x^2 + 5x + 3x^2 + 6x + 15 = x^3 + (2x^2 + 3x^2) + (5x + 6x) + 15.

Combine each group, $2x^2 + 3x^2 = 5x^2$ and $5x + 6x = 11x$ :

(x + 3)(x^2 + 2x + 5) = x^3 + 5x^2 + 11x + 15.

Worked example 5 Expand $(2x - 1)(3x^2 - 4x + 2)$

Distribute $2x$ across the trinomial, then $-1$ across it, carrying every sign:

2x(3x^2 - 4x + 2) = 6x^3 - 8x^2 + 4x,

-1(3x^2 - 4x + 2) = -3x^2 + 4x - 2.

Notice that $-1$ times $-4x$ is $+4x$ : multiplying two negatives gives a positive. Add the two rows and combine like terms:

6x^3 - 8x^2 + 4x - 3x^2 + 4x - 2 = 6x^3 - 11x^2 + 8x - 2.

The $x^2$ column gives $-8x^2 - 3x^2 = -11x^2$ , and the $x$ column gives $4x + 4x = 8x$ . The result is in standard form.

Check your understanding

Expand $(x + 1)(x^2 - 2x + 3)$ .

x^3 - x^2 + x + 3

x^3 + x^2 + x + 3

x^3 - 2x^2 + 3x + 3

x^3 - x^2 - x + 3

The degree of a product

Adding polynomials could sometimes lower the degree, when equal leading terms cancelled. Multiplication never springs that surprise, and the reason is short enough to prove in full.

Why the degree of a product is the sum of the degrees#

Write each polynomial in standard form, so its first term is its leading term. Suppose the first polynomial has leading term $a x^m$ and the second has leading term $b x^n$ . Here $a$ and $b$ are the leading coefficients, and $m$ and $n$ are the degrees. Among all the products you form, the one built from these two leading terms is

(a x^m)(b x^n) = ab\,x^{m+n}.

Every other product pairs a term of degree at most $m$ with a term of degree at most $n$ . In each such pair at least one factor has a strictly smaller degree, so every other product has degree strictly less than $m + n$ . That makes $ab\,x^{m+n}$ the single highest-degree term, with nothing else at its power to combine with or cancel it. Its coefficient $ab$ is the product of the two leading coefficients, and a leading coefficient is never zero. So $ab \neq 0$ , because a product of two nonzero numbers is nonzero. The top term genuinely survives, and the degree of the product is $m + n$ .

This is exactly where multiplication parts ways with addition. When you added polynomials, equal leading terms sat in the same column and could cancel if their coefficients were opposite, which sometimes dropped the degree. In a product the leading terms are multiplied, not added, and nonzero times nonzero is never zero, so a product cannot lose its leading term. The degree of a product is always the sum of the degrees, with no exceptions.

One more consequence deserves to be stated on its own. Multiply two polynomials and the result is always a polynomial again. Each partial product multiplies a real coefficient by a real coefficient, which is a real number. Each partial product also adds a whole-number exponent to a whole-number exponent, which is again a whole number. So every partial product is a coefficient times a nonnegative-integer power of $x$ , and a sum of such terms is a polynomial by definition. Mathematicians say the polynomials are closed under multiplication, the same way the whole numbers are closed under multiplication.

Check your understanding

What is the degree of the product $(4x^3 - x + 2)(x^2 + 5x - 1)$ ?

5

6

3

2

Common mistakes

Squaring a sum term by term. $(a + b)^2$ is not $a^2 + b^2$ . Squaring means $(a + b)(a + b)$ , which has four products and a middle term $2ab$ . Keeping only the first and last products and dropping the two cross terms is the most common error in the whole chapter. The same warning holds for any binomial times a binomial: you owe four products, not two.
Losing a sign on a negative term. Every term carries its sign into each product. In $(2x - 3)(x + 4)$ the product $(-3)(x)$ is $-3x$ and $(-3)(4)$ is $-12$ ; a single dropped minus flips a whole term and throws off the answer.
Forgetting to combine like terms. The four or six products are not the final answer. Gather the like powers and write standard form: $(x + 2)(x + 3)$ is $x^2 + 5x + 6$ , not $x^2 + 3x + 2x + 6$ left as it stands.
Multiplying the exponents instead of adding them. When you multiply powers of the same base you add the exponents: $x^2 \cdot x^3 = x^5$ , never $x^6$ . The coefficients multiply, but the exponents add.
Using FOIL where it does not apply. FOIL names only the binomial-times-binomial case. A binomial times a trinomial has six products, and forcing FOIL onto it quietly drops two of them. Fall back on the rule that always works, each term times each term.

Key takeaways

To multiply one term by one term, multiply the coefficients and add the exponents: $(3x^2)(4x^3) = 12x^5$ .
A single term distributes across every term of a polynomial: $a(b + c + d) = ab + ac + ad$ .
A binomial times a binomial is the distributive property used twice, giving four products: $(a + b)(c + d) = ac + ad + bc + bd$ . FOIL is only a nickname for this one case.
The general rule is each term of the first factor times each term of the second ( $m \times n$ products in all), then combine like terms and write standard form. The special products $(a + b)^2$ and $(a + b)(a - b)$ are this rule with the like terms already collected.
The degree of a product is the sum of the degrees, because the leading terms multiply and nonzero times nonzero cannot cancel. Contrast a sum, whose equal leading terms can cancel and drop the degree.
A product of polynomials is again a polynomial (closure).

Multiplying Polynomials

Learning goals

Multiplying one term by one term

Multiplying a polynomial by a single term

Multiplying two binomials

The general rule: every term times every term

The degree of a product

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)