Quick Reference
In this section we will learn how to multiply two polynomials.
| Method | Best used for | Core idea |
|---|---|---|
| Distributive property | Multiplication of any two polynomials | Multiply each term of the first by each term of the second |
| FOIL | Multiplication of two binomials only | First, Outer, Inner, Last |
| Long (vertical) method | Multiplication of three or more terms | Align partial products by degree in columns |
The Rule
To multiply two polynomials, apply the distributive property by multiplying each term of the first polynomial by each term of the second. Then use the exponent rule
$ax^{n}\cdot bx^{m}=ab\,x^{n+m}$and collect any like terms.
Product of Two Monomials
To multiply two monomials, multiply the coefficients and add the exponents:
$ (-4y^3)(5y^6) = (-4)(5)\,y^{3+6} = -20y^{9}. $Product of a Monomial and a Polynomial
Use the distributive property to multiply the monomial by each term of the polynomial:
$ 3x^2(2x^3 - 4x + 5) = 6x^5 - 12x^3 + 15x^2. $Product of Two Polynomials
Distribute each term of the first polynomial across all terms of the second, then collect like terms. For example, to expand $(3x-2)(2x^2-3x+4)$, distribute $3x$ and then $-2$:
$ \begin{aligned} 3x(2x^2-3x+4) &= 6x^3-9x^2+12x \\ -2(2x^2-3x+4) &= -4x^2+6x-8 \end{aligned} $Adding the results:
$ 6x^3+(-9-4)x^2+(12+6)x-8 = 6x^3-13x^2+18x-8. $
Example 1. Multiply $(4x^{2}-3x+5)(2x^{3}-x)$.
Solution
$ \begin{aligned} (4x^{2}-3x+5)(2x^{3}-x) &= 4x^{2}(2x^{3}-x)-3x(2x^{3}-x)+5(2x^{3}-x) \\ &= (8x^{5}-4x^{3})-(6x^{4}-3x^{2})+(10x^{3}-5x) \\ &= 8x^{5}-6x^{4}+(-4+10)x^{3}+3x^{2}-5x \\ &= 8x^{5}-6x^{4}+6x^{3}+3x^{2}-5x \end{aligned} $Degree of the Product
If $A$ is a polynomial of degree $n$ and $B$ is a polynomial of degree $m$, then the product $A\cdot B$ is a polynomial of degree $n+m$:
$\deg(A\cdot B)=\deg(A)+\deg(B).$In Example 1, $4x^{2}-3x+5$ has degree $2$ and $2x^{3}-x$ has degree $3$; the product has degree $2+3=5$, as expected.
Long Multiplication
When multiplying polynomials with three or more terms, a structured layout analogous to long multiplication of integers is often cleaner and less error-prone. The procedure is:
- Arrange both polynomials in descending order of degree. Write the one with more terms on top (the multiplicand).
- Form partial products: multiply the multiplicand by each term of the multiplier separately, writing each result on its own row.
- Align by degree: place each partial product so that terms of the same degree sit in the same column.
- Add column by column to obtain the final result.
Key tip: if the multiplicand is missing a term of some degree, that column is simply left blank (or filled with $0$) to keep the alignment correct.
Example 2. Multiply $(2x^3-x^2+5)$ by $(x^2+x-3)$.
Solution
Arrange in descending order and form the three partial products: $ \begin{aligned} x^2\cdot(2x^3-x^2+5) &= 2x^5-x^4+5x^2 \\ x\cdot(2x^3-x^2+5) &= 2x^4-x^3+5x \\ (-3)\cdot(2x^3-x^2+5) &= -6x^3+3x^2-15 \end{aligned} $ Add the partial products, aligning by degree: $ \begin{array}{rrrrrr} 2x^5 & -x^4 & & +5x^2 & & \\ & +2x^4 & -x^3 & & +5x & \\ & & -6x^3 & +3x^2 & & -15 \\ \hline 2x^5 & +x^4 & -7x^3 & +8x^2 & +5x & -15 \end{array} $ Therefore, $ (2x^3-x^2+5)(x^2+x-3) = 2x^5+x^4-7x^3+8x^2+5x-15. $Example 3. Multiply $(4x^2-3x+5)$ by $(2x^3-x)$ using the long multiplication method.
Solution
The multiplier $2x^3-x$ has no $x^2$ or constant term, so those columns are empty: $ \begin{aligned} 2x^3\cdot(4x^2-3x+5) &= 8x^5-6x^4+10x^3 \\ (-x)\cdot(4x^2-3x+5) &= -4x^3+3x^2-5x \end{aligned} $ Adding: $ \begin{array}{rrrrr} 8x^5 & -6x^4 & +10x^3 & & \\ & & -4x^3 & +3x^2 & -5x \\ \hline 8x^5 & -6x^4 & +6x^3 & +3x^2 & -5x \end{array} $ Therefore, $(4x^2-3x+5)(2x^3-x) = 8x^5-6x^4+6x^3+3x^2-5x$, confirming Example 1.FOIL Method for Multiplying Binomials
When multiplying two binomials, a helpful mnemonic called FOIL organizes the four products that arise from the distributive property. FOIL stands for First, Outer, Inner, Last.
Consider two binomials $(a+b)$ and $(c+d)$ where $a, b, c$ and $d$ represent any algebraic terms. For example, in (3x-2)(-2x^2+5x), we have $a=3x$, $b=-2$, $c=-2x^2$, and $d=5x$.The four FOIL products are:
- F (First): Multiply the first terms of each binomial: $a\cdot c=ac$.
- O (Outer): Multiply the outer terms: $a\cdot d=ad$.
- I (Inner): Multiply the inner terms: $b\cdot c=bc$.
- L (Last): Multiply the last terms: $b\cdot d=bd$.
Adding all four products: $(a+b)(c+d)=ac+ad+bc+bd$.

Example 4. Multiply $(x+2)(x-3)$ using the FOIL method.
Solution
- F (First): $x\cdot x = x^2$
- O (Outer): $x\cdot(-3) = -3x$
- I (Inner): $2\cdot x = 2x$
- L (Last): $2\cdot(-3) = -6$
Example 5. Multiply $(2y-5)(3y+1)$ using the FOIL method.
Solution
- F (First): $2y\cdot 3y = 6y^2$
- O (Outer): $2y\cdot 1 = 2y$
- I (Inner): $(-5)\cdot 3y = -15y$
- L (Last): $(-5)\cdot 1 = -5$
FOIL is a useful mnemonic for binomials, but it is simply the distributive property in disguise. The core rule is always to multiply every term in the first polynomial by every term in the second. For polynomials with more than two terms, FOIL does not directly apply; use the distributive property or the long multiplication method instead.
Determining Coefficients in the Product
Consider the product of a degree-3 polynomial by a degree-2 polynomial:
$ \begin{aligned} &(a_3x^3+a_2x^2+a_1x+a_0)(b_2x^2+b_1x+b_0) \\ &= a_3b_2\,x^5+(a_3b_1+a_2b_2)x^4+(a_3b_0+a_2b_1+a_1b_2)x^3 \\ &\quad+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x+a_0b_0. \end{aligned} $The product has degree $3+2=5$, and a clear pattern governs each coefficient:
The coefficient of $x^i$ in the product is the sum of all products $a_j b_k$ where $j+k=i$, with $j$ ranging over the degrees present in the first polynomial and $k$ over those in the second.
For example, to get the coefficient of $x^3$, collect all pairs $(j,k)$ with $j+k=3$: they are $(3,0)$, $(2,1)$, and $(1,2)$, giving $a_3b_0+a_2b_1+a_1b_2$.
This rule applies to the product of any two polynomials
$ (a_mx^m+\cdots+a_0)(b_nx^n+\cdots+b_0) $and provides an efficient way to find any single coefficient without expanding the entire product.
Example 6. Find the coefficient of $x^{100}$ in the product
$ (a_{75}x^{75}+a_{74}x^{74}+\cdots+a_0)(b_{60}x^{60}+b_{59}x^{59}+\cdots+b_0). $Solution
The degree of the product is $75+60=135$. For the term $x^{100}$, we need $j+k=100$ with $0\leq j\leq 75$ and $0\leq k\leq 60$. The largest possible $j$ is $75$ (giving $k=25$) and the smallest is $100-60=40$ (giving $k=60$). So $j$ ranges from $40$ to $75$, and the coefficient of $x^{100}$ is $ a_{75}b_{25}+a_{74}b_{26}+\cdots+a_{40}b_{60}. $ For comparison, the leading coefficient of the product (coefficient of $x^{135}$) has only one contributing pair, $j=75$ and $k=60$, so it equals $a_{75}b_{60}$.Example 7. Find the coefficient of $x^3$ in the product
$ (3x^4-2x^3+x^2-8x+7)(2x^3+5x^2+6x-3). $Solution
We need all pairs $(j,k)$ with $j+k=3$, where $0\leq j\leq 3$ and $0\leq k\leq 3$:| $j$ | $k$ | $a_j$ | $b_k$ | $a_jb_k$ |
|---|---|---|---|---|
| $3$ | $0$ | $-2$ | $-3$ | $6$ |
| $2$ | $1$ | $1$ | $6$ | $6$ |
| $1$ | $2$ | $-8$ | $5$ | $-40$ |
| $0$ | $3$ | $7$ | $2$ | $14$ |
Frequently Asked Questions