Multiplying Polynomials

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 of two nonzero polynomials to find their product's degree

Multiplying one term by one term

Before multiplying whole polynomials, start with the smallest piece: a single term times a single term. A term is a coefficient times a power of xx, 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:

(3x2)(4x3)=(3⋅4)(x2⋅x3).(3x^2)(4x^3) = (3 \cdot 4)(x^2 \cdot x^3).

The coefficients multiply the ordinary way, 3⋅4=123 \cdot 4 = 12. The powers combine by the product rule for exponents, which says xm⋅xn=xm+nx^m \cdot x^n = x^{m+n}, so you add the exponents:

(3x2)(4x3)=12 x2+3=12x5.(3x^2)(4x^3) = 12\,x^{2+3} = 12x^5.

Multiply the coefficients, add the exponents. You add rather than multiply the exponents because x2⋅x3x^2 \cdot x^3 is two xx‘s times three more xx‘s. That is five xx‘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:

(−5x4)(2x)=(−5⋅2)(x4⋅x1)=−10x5.(-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.a(b + c + d) = ab + ac + ad.

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

Worked example 1 Multiply 2x3(4x2−3x+5)2x^3(4x^2 - 3x + 5)

Distribute 2x32x^3 across all three terms, keeping each sign with its term:

2x3(4x2−3x+5)=(2x3)(4x2)+(2x3)(−3x)+(2x3)(5).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:

(2x3)(4x2)=8x5,(2x3)(−3x)=−6x4,(2x3)(5)=10x3.(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:

2x3(4x2−3x+5)=8x5−6x4+10x3.2x^3(4x^2 - 3x + 5) = 8x^5 - 6x^4 + 10x^3.

Check your understanding

Expand −3x2(2x3−x+4)-3x^2(2x^3 - x + 4).

Answer choices

Multiplying two binomials

Now both factors are sums, and the distributive property still does all the work. Look at (x+2)(x+3)(x + 2)(x + 3). If xx is a positive length, then x+2x + 2 and x+3x + 3 are side lengths, and the product 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 partial products.

Area model for (x + 2)(x + 3)A large rectangle partitioned into a two-by-two grid of areas: top-left x squared, top-right 3x, bottom-left 2x, bottom-right 6.x3x2x²3x2x6
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, 3x3x on the top right and 2x2x on the bottom left, are the Outer and Inner products. They carry the same power of xx, 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)(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⋅x+x⋅3+2⋅x+2⋅3=x2+3x+2x+6.(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:

x2+3x+2x+6=x2+5x+6.x^2 + 3x + 2x + 6 = x^2 + 5x + 6.

These are the four areas from the figure, x2x^2, 3x3x, 2x2x, and 66, added together. Leaving the answer as x2+3x+2x+6x^2 + 3x + 2x + 6 is not wrong, only unfinished; combining like terms is what puts it in standard form.

The same thing happens for any two binomials, not just this one. Hold the second binomial (c+d)(c + d) together as a single block for a moment: (a+b)(a + b) distributes across it, and each of the two pieces distributes again.

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

Distribute a+ba + b over c+dc + d exactly as the distributive property allows, then distribute each of the two pieces in turn:

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

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, the same idea that runs the general rule below.

Those four products have a well-known nickname. Reading them in the order First, Outer, Inner, Last spells FOIL: the product of the First terms (acac), the Outer terms (adad), the Inner terms (bcbc), and the Last terms (bdbd). FOIL is a handy label, but it only counts to four. It says nothing about a binomial times a trinomial, where there are six products instead. The rule that always works, no matter how many terms each factor has, is the plain idea behind FOIL: each term times each term.

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)(2x - 3)(x + 4)

The four products, with signs kept in place, are

(2x)(x)+(2x)(4)+(−3)(x)+(−3)(4)=2x2+8x−3x−12.(2x)(x) + (2x)(4) + (-3)(x) + (-3)(4) = 2x^2 + 8x - 3x - 12.

Combine the two middle terms, 8x−3x=5x8x - 3x = 5x:

(2x−3)(x+4)=2x2+5x−12.(2x - 3)(x + 4) = 2x^2 + 5x - 12.

The term (−3)(x)(-3)(x) is −3x-3x and (−3)(4)(-3)(4) is −12-12. 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 - 4)(x + 6).

Answer choices

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 mm terms times a factor with nn terms produces m×nm \times n products before you combine anything. A binomial times a binomial gives 2×2=42 \times 2 = 4 products, which is the case FOIL covers; a binomial times a trinomial gives 2×3=62 \times 3 = 6.

The special products from the factorizations chapter are this same rule, just with the like terms already collected. Squaring (a+b)(a+b)(a + b)(a + b) gives four products that collapse to a2+2ab+b2a^2 + 2ab + b^2, and the difference of squares (a+b)(a−b)(a + b)(a - b) has its two cross terms cancel to leave a2−b2a^2 - b^2. Neither is a separate fact to memorize; this lesson keeps the method general instead.

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.

Grid method for (x + 3)(x squared + 2x + 5)A two-by-three grid. Row headers are x and 3; column headers are x squared, 2x, and 5. The cells are x cubed, 2x squared, 5x, 3x squared, 6x, and 15.×x²2x5x3x³2x²5x3x²6x15
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)(x2+2x+5)(x + 3)(x^2 + 2x + 5)

The binomial has two terms and the trinomial has three, so expect 2×3=62 \times 3 = 6 products. Distribute xx across the trinomial, then 33 across it:

x(x2+2x+5)=x3+2x2+5x,3(x2+2x+5)=3x2+6x+15.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:

x3+2x2+5x+3x2+6x+15=x3+(2x2+3x2)+(5x+6x)+15.x^3 + 2x^2 + 5x + 3x^2 + 6x + 15 = x^3 + (2x^2 + 3x^2) + (5x + 6x) + 15.

Combine each group, 2x2+3x2=5x22x^2 + 3x^2 = 5x^2 and 5x+6x=11x5x + 6x = 11x:

(x+3)(x2+2x+5)=x3+5x2+11x+15.(x + 3)(x^2 + 2x + 5) = x^3 + 5x^2 + 11x + 15.

Worked example 5 Expand (2x−1)(3x2−4x+2)(2x - 1)(3x^2 - 4x + 2)

Distribute 2x2x across the trinomial, then −1-1 across it, carrying every sign:

2x(3x2−4x+2)=6x3−8x2+4x,2x(3x^2 - 4x + 2) = 6x^3 - 8x^2 + 4x,−1(3x2−4x+2)=−3x2+4x−2.-1(3x^2 - 4x + 2) = -3x^2 + 4x - 2.

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

6x3−8x2+4x−3x2+4x−2=6x3−11x2+8x−2.6x^3 - 8x^2 + 4x - 3x^2 + 4x - 2 = 6x^3 - 11x^2 + 8x - 2.

The x2x^2 column gives −8x2−3x2=−11x2-8x^2 - 3x^2 = -11x^2, and the xx column gives 4x+4x=8x4x + 4x = 8x. The result is in standard form.

Check your understanding

Expand (x+1)(x2−2x+3)(x + 1)(x^2 - 2x + 3).

Answer choices

The degree of a product

Adding polynomials could sometimes lower the degree, when equal leading terms canceled. For two nonzero polynomials, multiplying them never springs that surprise. Multiply the two leading terms, axmax^m from one polynomial and bxnbx^n from the other, and you get ab xm+nab\,x^{m+n}. Every other partial product pairs a lower-degree term from one factor with a term from the other, so its power of xx is strictly below m+nm + n, nothing left to cancel the top term. And abab itself cannot be zero, because aa and bb are leading coefficients, and a leading coefficient is never zero. So the degree of the product is always m+nm + n, the sum of the two degrees. Multiplication cannot lose its leading term the way addition sometimes can. One more fact falls out for free: multiplying two polynomials always gives another polynomial, never anything of a different shape.

Check your understanding

What is the degree of the product (4x3−x+2)(x2+5x−1)(4x^3 - x + 2)(x^2 + 5x - 1)?

Answer choices

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

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Why the top term always survives

Write each polynomial in standard form, so its first term is its leading term. Suppose the first polynomial has leading term axmax^m and the second has leading term bxnbx^n, where aa and bb are the leading coefficients and mm and nn are the degrees. Among all the products you form, the one built from these two leading terms is

(axm)(bxn)=ab xm+n.(ax^m)(bx^n) = ab\,x^{m+n}.

Every other product pairs a term of degree at most mm with a term of degree at most nn, and in each such pair at least one factor has a strictly smaller degree. So every other product has degree strictly less than m+nm + n, and ab xm+nab\,x^{m+n} is the single highest-degree term, with nothing else at its power to cancel it. Its coefficient abab is a product of two nonzero numbers, since a leading coefficient is never zero, so ab≠0ab \neq 0 and the top term genuinely survives.

Polynomials are closed under multiplication

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, and 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 xx, 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.

A bit of history (optional)

Teachers kept meeting the same wrong answer. A student would multiply two binomials, write down the first product and the last, and lose the two in the middle. First, Outer, Inner, Last gives FOIL, a mnemonic that was appearing in American algebra textbooks by the late 1920s. It spread because it is easy to chant and hard to forget, which is what a mnemonic is for.

It has also drawn steady complaints, and the complaints are fair: FOIL can only count to four, and it names nothing about why the four products exist, only how to remember them. Keep the chant if it helps, but remember the rule underneath it, every term of the first factor times every term of the second.