Multiply Polynomials

By the end of this section, you will be able to: multiply monomials, multiply a polynomial by a monomial, multiply a binomial by a binomial, multiply a polynomial by a polynomial, multiply special products, and multiply polynomial functions.

Multiply monomials

We are ready to perform operations on polynomials. Since monomials are algebraic expressions, we can use the properties of exponents to multiply monomials.

Example. Multiply: (a) $\left(3x^2\right)\left(-4x^3\right)$ and (b) $\left(\tfrac{5}{6}x^3y\right)\left(12xy^2\right)$ .

For part (a):

\begin{array}{lrcl} \text{Use the Commutative Property to rearrange the terms.} & \left(3x^2\right)\left(-4x^3\right) &=& 3\cdot(-4)\cdot x^2\cdot x^3 \\[4pt] \text{Multiply.} & &=& -12x^5 \end{array}

For part (b):

\begin{array}{lrcl} \text{Use the Commutative Property to rearrange the terms.} & \left(\tfrac{5}{6}x^3y\right)\left(12xy^2\right) &=& \tfrac{5}{6}\cdot 12\cdot x^3\cdot x\cdot y\cdot y^2 \\[4pt] \text{Multiply.} & &=& 10x^4y^3 \end{array}

Multiply: $\left(5y^7\right)\left(-7y^4\right)$

Multiply: $\left(\tfrac{2}{5}a^4b^3\right)\left(15ab^3\right)$

Multiply a polynomial by a monomial

Multiplying a polynomial by a monomial is really just applying the Distributive Property.

Example. Multiply: (a) $-2y\left(4y^2+3y-5\right)$ and (b) $3x^3y\left(x^2-8xy+y^2\right)$ .

For part (a):

\begin{array}{lrcl} \text{Distribute.} & -2y\left(4y^2+3y-5\right) &=& -2y\cdot 4y^2+(-2y)\cdot 3y-(-2y)\cdot 5 \\[4pt] \text{Multiply.} & &=& -8y^3-6y^2+10y \end{array}

For part (b):

\begin{array}{lrcl} \text{Distribute.} & 3x^3y\left(x^2-8xy+y^2\right) &=& 3x^3y\cdot x^2+\left(3x^3y\right)(-8xy)+\left(3x^3y\right)\cdot y^2 \\[4pt] \text{Multiply.} & &=& 3x^5y-24x^4y^2+3x^3y^3 \end{array}

Multiply: $-3y\left(5y^2+8y-7\right)$

Multiply: $4x^2\left(2x^2-3x+5\right)$

Multiply a binomial by a binomial

Just as there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

Example. Multiply $(y+5)(y+8)$ .

\begin{array}{lrcl} \text{Distribute } (y+8). & (y+5)(y+8) &=& y(y+8)+5(y+8) \\[4pt] \text{Distribute again.} & &=& y^2+8y+5y+40 \\[4pt] \text{Combine like terms.} & &=& y^2+13y+40 \end{array}

Multiply: $(x+8)(x+9)$

Multiply: $(3c+4)(5c-2)$

The FOIL method

If you multiply binomials often enough you may notice a pattern. Notice that the first term in the result is the product of the first terms in each binomial. The second and third terms are the product of multiplying the two outer terms and then the two inner terms. And the last term results from multiplying the two last terms.

We abbreviate “First, Outer, Inner, Last” as FOIL. Using FOIL as another method of multiplying binomials ensures we find all four products. For example, multiplying $(x+3)(x+7)$ :

\begin{array}{lrcl} \text{First, Outer, Inner, Last.} & (x+3)(x+7) &=& x^2+7x+3x+21 \\[4pt] \text{Combine like terms.} & &=& x^2+10x+21 \end{array}

The FOIL method only applies to multiplying two binomials, not other polynomials!

Use the FOIL method to multiply two binomials. For

(a+b)(c+d)

: Step 1. Multiply the First terms

a\cdot c

. Step 2. Multiply the Outer terms

a\cdot d

. Step 3. Multiply the Inner terms

b\cdot c

. Step 4. Multiply the Last terms

b\cdot d

. Step 5. Combine like terms, when possible.

Example. Multiply using FOIL: (a) $(y-7)(y+4)$ and (b) $(4x+3)(2x-5)$ .

For part (a):

\begin{array}{lrcl} \text{Multiply First, Outer, Inner, Last.} & (y-7)(y+4) &=& y^2+4y-7y-28 \\[4pt] \text{Combine like terms.} & &=& y^2-3y-28 \end{array}

For part (b):

\begin{array}{lrcl} \text{Multiply First, Outer, Inner, Last.} & (4x+3)(2x-5) &=& 8x^2-20x+6x-15 \\[4pt] \text{Combine like terms.} & &=& 8x^2-14x-15 \end{array}

Multiply using FOIL: $(x-7)(x+5)$

Multiply using FOIL: $(3x+7)(5x-2)$

The final products in the last example were trinomials because we could combine the two middle terms. This is not always the case.

Example. Multiply: (a) $\left(n^2+4\right)(n-1)$ and (b) $(3pq+5)(6pq-11)$ .

For part (a):

\begin{array}{lrcl} \text{Multiply First, Outer, Inner, Last.} & \left(n^2+4\right)(n-1) &=& n^3-n^2+4n-4 \\[4pt] \text{Combine like terms — there are none.} & &=& n^3-n^2+4n-4 \end{array}

For part (b):

\begin{array}{lrcl} \text{Multiply First, Outer, Inner, Last.} & (3pq+5)(6pq-11) &=& 18p^2q^2-33pq+30pq-55 \\[4pt] \text{Combine like terms.} & &=& 18p^2q^2-3pq-55 \end{array}

Multiply: $\left(x^2+6\right)(x-8)$

Multiply: $(2ab+5)(4ab-4)$

The Vertical Method

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers, lining up partial products in columns and adding.

Example. Multiply using the Vertical Method: $(3y-1)(2y-6)$ .

It does not matter which binomial goes on the top. Multiply $3y-1$ by $-6$ , then multiply $3y-1$ by $2y$ , and add like terms:

\begin{array}{r} 3y-1 \\ \times\phantom{0}2y-6 \\ \hline -18y+6 \\ 6y^2-\phantom{0}2y\phantom{{}+6} \\ \hline 6y^2-20y+6 \end{array}

Notice the partial products are the same as the terms in the FOIL method.

Multiply using the Vertical Method: $(5m-7)(3m-6)$

Multiply using the Vertical Method: $(6b-5)(7b-3)$

Multiplying two binomials. To multiply binomials, use the: Distributive Property, FOIL Method, or Vertical Method.

Multiply a polynomial by a polynomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we are ready to multiply a polynomial by a polynomial. Remember, FOIL will not work in this case, but we can use either the Distributive Property or the Vertical Method.

Example. Multiply $(b+3)\left(2b^2-5b+8\right)$ using the Distributive Property.

\begin{array}{lrcl} \text{Distribute.} & (b+3)\left(2b^2-5b+8\right) &=& b\left(2b^2-5b+8\right)+3\left(2b^2-5b+8\right) \\[4pt] \text{Multiply.} & &=& 2b^3-5b^2+8b+6b^2-15b+24 \\[4pt] \text{Combine like terms.} & &=& 2b^3+b^2-7b+24 \end{array}

Using the Vertical Method, it is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products that way:

\begin{array}{r} 2b^2-\phantom{0}5b+8 \\ \times\phantom{2b^2-{}}b+3 \\ \hline 6b^2-15b+24 \\ 2b^3-\phantom{0}5b^2+\phantom{0}8b\phantom{{}+24} \\ \hline 2b^3+\phantom{0}b^2-\phantom{0}7b+24 \end{array}

Multiply $(y-3)\left(y^2-5y+2\right)$ .

Multiply $(x+4)\left(2x^2-3x+5\right)$ .

Multiplying a polynomial by a polynomial. To multiply a trinomial by a binomial, use the: Distributive Property or Vertical Method.

Multiply special products

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and multiplying them, there is less work to do if you learn to use a pattern.

Look at these results, where each squared binomial produces a trinomial:

\begin{array}{l} (x+9)^2=(x+9)(x+9)=x^2+9x+9x+81=x^2+18x+81 \\[4pt] (y-7)^2=(y-7)(y-7)=y^2-7y-7y+49=y^2-14y+49 \\[4pt] (2x+3)^2=(2x+3)(2x+3)=4x^2+6x+6x+9=4x^2+12x+9 \end{array}

The first term is the square of the first term of the binomial, the last term is the square of the last term, and the middle term is double the product of the two terms. This gives the Binomial Squares Pattern.

Binomial Squares Pattern. If $a$ and $b$ are real numbers,

(a+b)^2=a^2+2ab+b^2 \qquad (a-b)^2=a^2-2ab+b^2.

To square a binomial, square the first term, square the last term, and double their product.

Example. Multiply: (a) $(x+5)^2$ and (b) $(2x-3y)^2$ .

For part (a), using $(a+b)^2=a^2+2ab+b^2$ :

\begin{array}{lrcl} \text{Use the pattern.} & (x+5)^2 &=& x^2+2\cdot x\cdot 5+5^2 \\[4pt] \text{Simplify.} & &=& x^2+10x+25 \end{array}

For part (b), using $(a-b)^2=a^2-2ab+b^2$ :

\begin{array}{lrcl} \text{Use the pattern.} & (2x-3y)^2 &=& (2x)^2-2\cdot 2x\cdot 3y+(3y)^2 \\[4pt] \text{Simplify.} & &=& 4x^2-12xy+9y^2 \end{array}

Multiply: $(x+9)^2$

Multiply: $(2c-d)^2$

We just saw a pattern for squaring binomials. Similarly, there is a pattern for another product of binomials. A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference, is called a conjugate pair and is of the form $(a-b),\ (a+b)$ .

Conjugate pair. A conjugate pair is two binomials of the form

(a-b),\ (a+b)

. The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

There is a nice pattern for finding the product of conjugates. Notice the two middle terms you get from FOIL combine to $0$ in every case, leaving a product of the form $a^2-b^2$ , called a difference of squares.

Product of Conjugates Pattern. If $a$ and $b$ are real numbers,

(a-b)(a+b)=a^2-b^2.

The product is called a difference of squares. To multiply conjugates, square the first term, square the last term, and write it as a difference of squares.

Example. Multiply using the product of conjugates pattern: (a) $(2x+5)(2x-5)$ and (b) $(5m-9n)(5m+9n)$ .

For part (a):

\begin{array}{lrcl} \text{Use the pattern.} & (2x+5)(2x-5) &=& (2x)^2-5^2 \\[4pt] \text{Simplify.} & &=& 4x^2-25 \end{array}

For part (b):

\begin{array}{lrcl} \text{Use the pattern.} & (5m-9n)(5m+9n) &=& (5m)^2-(9n)^2 \\[4pt] \text{Simplify.} & &=& 25m^2-81n^2 \end{array}

Multiply: $(6x+5)(6x-5)$

Multiply: $(4p-7q)(4p+7q)$

The special product patterns look similar, so it is important to recognize when to use each. Squaring a binomial gives a trinomial whose middle term is double the product of the terms; multiplying conjugates gives a binomial — a difference of squares with no middle term.

Example. Choose the appropriate pattern and use it to find the product: (a) $(2x-3)(2x+3)$ , (b) $(8x-5)^2$ , (c) $(6m+7)^2$ , and (d) $(5x-6)(6x+5)$ .

For part (a), these are conjugates, so it fits the Product of Conjugates pattern:

\begin{array}{lrcl} \text{Use the pattern.} & (2x-3)(2x+3) &=& (2x)^2-3^2 \\[4pt] \text{Simplify.} & &=& 4x^2-9 \end{array}

For part (b), we square a binomial, so it fits the Binomial Squares pattern:

\begin{array}{lrcl} \text{Use the pattern.} & (8x-5)^2 &=& (8x)^2-2\cdot 8x\cdot 5+5^2 \\[4pt] \text{Simplify.} & &=& 64x^2-80x+25 \end{array}

For part (c), again a binomial square:

\begin{array}{lrcl} \text{Use the pattern.} & (6m+7)^2 &=& (6m)^2+2\cdot 6m\cdot 7+7^2 \\[4pt] \text{Simplify.} & &=& 36m^2+84m+49 \end{array}

For part (d), this product does not fit the patterns, so we use FOIL:

\begin{array}{lrcl} \text{Use FOIL.} & (5x-6)(6x+5) &=& 30x^2+25x-36x-30 \\[4pt] \text{Simplify.} & &=& 30x^2-11x-30 \end{array}

Choose the appropriate pattern and find the product: $(9b-2)(2b+9)$

Choose the appropriate pattern and find the product: $(9p-4)^2$

Choose the appropriate pattern and find the product: $(7y+1)^2$

Multiply polynomial functions

Just as polynomials can be multiplied, polynomial functions can also be multiplied.

Multiplication of Polynomial Functions. For functions

f(x)

and

g(x)

(f\cdot g)(x)=f(x)\cdot g(x)

Example. For functions $f(x)=x+2$ and $g(x)=x^2-3x-4$ , find (a) $(f\cdot g)(x)$ and (b) $(f\cdot g)(2)$ .

For part (a):

\begin{array}{lrcl} \text{Substitute for } f(x) \text{ and } g(x). & (f\cdot g)(x) &=& (x+2)\left(x^2-3x-4\right) \\[4pt] \text{Multiply the polynomials.} & &=& x\left(x^2-3x-4\right)+2\left(x^2-3x-4\right) \\[4pt] \text{Distribute.} & &=& x^3-3x^2-4x+2x^2-6x-8 \\[4pt] \text{Combine like terms.} & &=& x^3-x^2-10x-8 \end{array}

For part (b), we found $(f\cdot g)(x)$ above, so we substitute $x=2$ :

\begin{array}{lrcl} \text{Substitute } x=2. & (f\cdot g)(2) &=& 2^3-2^2-10\cdot 2-8 \\[4pt] \text{Simplify.} & &=& 8-4-20-8 \\[4pt] & &=& -24 \end{array}

For $f(x)=x-5$ and $g(x)=x^2-2x+3$ , find $(f\cdot g)(x)$ .

For $f(x)=x-5$ and $g(x)=x^2-2x+3$ , find $(f\cdot g)(2)$ .

Key terms

FOIL — a method for multiplying two binomials by adding the products of the First, Outer, Inner, and Last terms. Vertical Method — a method for multiplying polynomials that lines up partial products in columns, like whole number multiplication. conjugate pair — two binomials of the form $(a-b),\ (a+b)$ with the same first and last terms, one a sum and one a difference. binomial squares pattern — $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ . product of conjugates pattern — $(a-b)(a+b)=a^2-b^2$ . difference of squares — a binomial of the form $a^2-b^2$ , the product of a conjugate pair.

This section is adapted from Intermediate Algebra 2e, Section 5.3: Multiply Polynomials by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the Be Prepared quiz, media links, and end-of-section exercises; recreated the FOIL and Vertical Method worked examples as aligned equation blocks; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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