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Multiplying binomials by polynomials review

A review of how to multiply binomials like 1 + x by polynomials with more than two terms like x^2 - 5x - 6
We already know how to multiply binomials like (x+2)(x7). In this article, we review a slightly more complicated skill: multiplying binomials by polynomials with more than two terms.

Example

Expand and simplify.
(1+x)(x25x6)
Apply the distributive property.
(1+x)(x25x6)=1(x25x6)+x(x25x6)
Apply the distributive property again.
=1(x2)+1(5x)+1(6)+x(x2)+x(5x)+x(6)
Notice the pattern. We multiplied each term in the binomial by each term in the trinomial.
Simplify.
=x25x6+x35x26x=x34x211x6
Want to learn more about multiplying binomials and polynomials? Check out this video.
Practice
Expand and simplify.
(c26)(2c2+3c1)

Want more practice? Check out this exercise.

Want to join the conversation?

  • piceratops seed style avatar for user ltoutai
    How do you simplify an equation
    (12 votes)
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  • duskpin seedling style avatar for user JeicobDaBoss
    What would be considered the fastest method of multiplying long polynomials?
    (5 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      Great question! Yes there is a shortcut, which efficiently combines all the terms for each power of x.

      If only one variable, say x, appears in the two polynomials to be multiplied together, then you can use a technique similar to the Vedic multiplication arithmetic technique called vertical and crosswise (which you can look up online to get the idea). Write the terms of each polynomial in order of descending powers of x. For any missing exponents on x, use a coefficient of zero.

      Example: multiply (6x^4 + 3x^3 - 5x + 7) by (2x^3 - 4x^2 - x + 8).

      Write
      (6x^4 + 3x^3 + 0x^2 - 5x + 7)
      (0x^4 + 2x^3 - 4x^2 - x + 8).

      These two polynomials are now each written with five coefficients.

      The idea is to multiply the first coefficient by the first coefficient, then cross multiply the first two coefficients by the first two coefficients, then the first three by the first three, then the first four by the first four, then the first five by the first five, then the last four by the last four, then the last three by the last three, then the last two by the last two, then finally the last one by the last one.

      There are no terms of degree 9 or higher.
      Coefficient of x^8 is 6(0) = 0.
      Coefficient of x^7 is 6(2) + 3(0) = 12.
      Coefficient of x^6 is 6(-4) + 3(2) + 0(0) = -18.
      Coefficient of x^5 is 6(-1) + 3(-4) + 0(2) + (-5)(0) = -18.
      Coefficient of x^4 is 6(8) + 3(-1) + 0(-4) + (-5)(2) + 7(0) = 35.
      Coefficient of x^3 is 3(8) + 0(-1) + (-5)(-4) + 7(2) = 58.
      Coefficient of x^2 is 0(8) + (-5)(-1) + 7(-4) = -23.
      Coefficient of x is -5(8) + 7(-1) = -47.
      Constant term (coefficient on x^0) is 7(8) = 56.

      The answer is
      12x^7 - 18x^6 - 18x^5 + 35x^4 + 58x^3 - 23x^2 - 47x + 56.

      Have a blessed, wonderful day!
      (52 votes)
  • marcimus pink style avatar for user Sherry
    So, you can apply the distributive property to one of these equations, or you can use the graph/colored boxes thing? Which way is more efficient?
    (8 votes)
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    • mr pink green style avatar for user David Severin
      There is FOIL, double distribution, and box method as 3 ways of doing the exact same thing, they all give the same 4 terms with the middle terms usually combined. FOIL is probably less efficient because it is limited to multiplying two binomials. The box method may require more writing, but might make more sense in exactly what is going on. Double distribution and box method can easily be expanded to larger polynomials in the future. Efficient may be in the eye of the beholder, so find which one works best for you. Many teachers will require you to learn all so that you can make an educated decision which is best, and if you do it enough and have a pretty good math brain, you will be able to do it in your head.
      (15 votes)
  • blobby green style avatar for user Taha iftikhar
    in the vid they never clearly said which goes first they just said its the only one cubed and do I do y squared first or y alone first
    (6 votes)
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    • stelly blue style avatar for user Kim Seidel
      Generally, you should write your answer in standard form. This would have the term with y^2 first, then y, then the constant. However, it is usually not required unless the instructions specify your answer must be in standard form.
      (12 votes)
  • duskpin tree style avatar for user WolfyNightshade
    Am I ever going to need this type of math in the real world?
    (7 votes)
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  • aqualine tree style avatar for user amankeshwani
    is this apart of the algebra 2 course
    (4 votes)
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  • aqualine ultimate style avatar for user Korbyn Vik
    What do you think is the best way to multiply and simplify a problem?
    (2 votes)
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  • blobby green style avatar for user prasannaa2020
    Most Confusing thing ever!!
    (7 votes)
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  • blobby green style avatar for user Ivan Wilkie
    when you are multiplying without a coefficient how do you deal with the exponent.
    (2 votes)
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  • piceratops sapling style avatar for user chloe cartwright
    for those who are having trouble with the distributive method you should try the box method its layed out in a way that may end up being easier to understand. its hard to put that method on the the chat but i hope that info helps.(i would look it up if you don´t understand what i mean by box method)
    (6 votes)
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