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9.2: Multiplication of Polynomials

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Multiply a polynomial by a monomial.
  • Multiply a polynomial by a binomial.
  • Solve problems using multiplication of polynomials.


Just as we can add and subtract polynomials, we can also multiply them. The Distributive Property and the techniques you’ve learned for dealing with exponents will be useful here.

Multiplying a Polynomial by a Monomial

When multiplying polynomials, we must remember the exponent rules that we learned in the last chapter. Especially important is the product rule: \begin{align*}x^n \cdot x^m=x^{n+m}\end{align*}xnxm=xn+m.

If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any number and we apply the product rule on each variable separately.

Example 1

Multiply the following monomials.

a) \begin{align*}(2x^2)(5x^3)\end{align*}(2x2)(5x3)

b) \begin{align*}(-3y^4)(2y^2)\end{align*}(3y4)(2y2)

c) \begin{align*}(3xy^5)(-6x^4y^2)\end{align*}(3xy5)(6x4y2)

d) \begin{align*}(-12a^2b^3c^4)(-3a^2b^2)\end{align*}(12a2b3c4)(3a2b2)


a) \begin{align*}(2x^2)(5x^3)=(2 \cdot 5) \cdot (x^2 \cdot x^3)=10x^{2+3} = 10x^5\end{align*}(2x2)(5x3)=(25)(x2x3)=10x2+3=10x5

b) \begin{align*}(-3y^4)(2y^2)=(-3 \cdot 2) \cdot (y^4 \cdot y^2)=-6y^{4+2}=-6y^6\end{align*}(3y4)(2y2)=(32)(y4y2)=6y4+2=6y6

c) \begin{align*}(3xy^5)(-6x^4y^2)=-18x^{1+4}y^{5+2}=-18x^5y^7\end{align*}(3xy5)(6x4y2)=18x1+4y5+2=18x5y7

d) \begin{align*}(-12a^2b^3c^4)(-3a^2b^2)=36a^{2+2}b^{3+2}c^4 = 36a^4b^5c^4\end{align*}(12a2b3c4)(3a2b2)=36a2+2b3+2c4=36a4b5c4

To multiply a polynomial by a monomial, we have to use the Distributive Property. Remember, that property says that \begin{align*}a(b + c) = ab + ac\end{align*}a(b+c)=ab+ac.

Example 2


a) \begin{align*}3(x^2+3x-5)\end{align*}3(x2+3x5)

b) \begin{align*}4x(3x^2-7)\end{align*}4x(3x27)

c) \begin{align*}-7y(4y^2-2y+1)\end{align*}7y(4y22y+1)


a) \begin{align*}3(x^2+3x-5)=3(x^2)+3(3x)-3(5)=3x^2+9x-15\end{align*}3(x2+3x5)=3(x2)+3(3x)3(5)=3x2+9x15

b) \begin{align*}4x(3x^2-7)=(4x)(3x^2)+(4x)(-7)=12x^3-28x\end{align*}4x(3x27)=(4x)(3x2)+(4x)(7)=12x328x

c) \begin{align*}-7y(4y^2-2y+1)=(-7y)(4y^2)+(-7y)(-2y)+(-7y)(1)=-28y^3+14y^2-7y\end{align*}7y(4y22y+1)=(7y)(4y2)+(7y)(2y)+(7y)(1)=28y3+14y27y

Notice that when we use the Distributive Property, the problem becomes a matter of just multiplying monomials by monomials and adding all the separate parts together.

Example 3


a) \begin{align*}2x^3(-3x^4+2x^3-10x^2+7x+9)\end{align*}

b) \begin{align*}-7a^2bc^3(5a^2-3b^2-9c^2)\end{align*}


a) \begin{align*}2x^3(-3x^4+2x^3-10x^2+7x+9)&=(2x^3)(-3x^4)+(2x^3)(2x^3)+(2x^3)(-10x^2)+(2x^3)(7x)+(2x^3)(9)\\ & = -6x^7+4x^6-20x^5+14x^4+18x^3\end{align*}

b) \begin{align*}-7a^2bc^3(5a^2-3b^2-9c^2) & = (-7a^2bc^3)(5a^2)+(-7a^2bc^3)(-3b^2)+(-7a^2bc^3)(-9c^2)\\ & = -35a^4bc^3 + 21a^2b^3c^3 + 63a^2bc^5\end{align*}

Multiplying Two Polynomials

Let’s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of two binomials will take the form \begin{align*}(a+b)(c+d)\end{align*}.

We can still use the Distributive Property here if we do it cleverly. First, let’s think of the first set of parentheses as one term. The Distributive Property says that we can multiply that term by \begin{align*}c\end{align*}, multiply it by \begin{align*}d\end{align*}, and then add those two products together: \begin{align*}(a+b)(c+d)=(a+b) \cdot c+(a+b) \cdot d\end{align*}.

We can rewrite this expression as \begin{align*}c(a+b)+d(a+b)\end{align*}. Now let’s look at each half separately. We can apply the distributive property again to each set of parentheses in turn, and that gives us \begin{align*}c(a+b)+d(a+b)=ca+cb+da+db\end{align*}.

What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial.

Example 4

Multiply and simplify: \begin{align*}(2x+1)(x+3)\end{align*}


We must multiply each term in the first polynomial by each term in the second polynomial. Let’s try to be systematic to make sure that we get all the products.

First, multiply the first term in the first set of parentheses by all the terms in the second set of parentheses.

Now we’re done with the first term. Next we multiply the second term in the first parenthesis by all terms in the second parenthesis and add them to the previous terms.

Now we’re done with the multiplication and we can simplify:


This way of multiplying polynomials is called in-line multiplication or horizontal multiplication. Another method for multiplying polynomials is to use vertical multiplication, similar to the vertical multiplication you learned with regular numbers.

Example 5

Multiply and simplify:

a) \begin{align*}(4x-5)(x-20)\end{align*}

b) \begin{align*}(3x-2)(3x+2)\end{align*}

c) \begin{align*}(3x^2+2x-5)(2x-3)\end{align*}

d) \begin{align*}(x^2-9)(4x^4+5x^2-2)\end{align*}


a) With horizontal multiplication this would be


To do vertical multiplication instead, we arrange the polynomials on top of each other with like terms in the same columns:

\begin{align*}& \qquad \quad \ 4x-5\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\; x-20\;\;}\\ & \quad \ -80x+100\\ & \underline{4x^2-5x \;\;\;\;\;\;\;\;\;\;\;}\\ & 4x^2-85x+100\end{align*}

Both techniques result in the same answer: \begin{align*}4x^2-85x+100\end{align*}. We’ll use vertical multiplication for the other problems.

b) \begin{align*}& \qquad \ \ 3x-2\\ & \underline{\;\;\;\;\;\;\;\;\; 3x+2}\\ & \qquad \ \ 6x-4\\ & \underline{9x^2-6x \;\;\;\;\;\;}\\ & 9x^2+0x-4\end{align*}

The answer is \begin{align*}9x^2-4\end{align*}.

c) It’s better to place the smaller polynomial on the bottom:

\begin{align*}& \qquad \quad 3x^2+2x-5\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x-3\;\;\;}\\ & \quad \ \ -9x^2-6x+15\\ & \underline{6x^3+4x^2-10x \;\;\;\;\;\;\;\;}\\ & 6x^3-5x^2-16x+15\end{align*}

The answer is \begin{align*}6x^3-5x^2-16x+15\end{align*}.

d) Set up the multiplication vertically and leave gaps for missing powers of \begin{align*}x\end{align*}:

\begin{align*}& \qquad \quad \ \ \ 4x^4+5x^2-2\\ &\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x^2-9}\\ & \quad \ \ -36x^4-45x^2+18\\ & \underline{4x^6+\;5x^4\;-\;2x^2 \;\;\;\;\;\;\;\;\;}\\ & 4x^6-31x^4-47x^2+18\end{align*}

The answer is \begin{align*}4x^6-31x^4-47x^2+18\end{align*}.

The Khan Academy video at http://www.youtube.com/watch?v=Sc0e6xrRJYY shows how multiplying two binomials together is related to the distributive property.

Solve Real-World Problems Using Multiplication of Polynomials

In this section, we’ll see how multiplication of polynomials is applied to finding the areas and volumes of geometric shapes.

Example 6

Find the areas of the following figures:



Find the volumes of the following figures:




a) We use the formula for the area of a rectangle: \begin{align*}\text{Area} = \text{length} \times \text{width}\end{align*}.

For the big rectangle:

\begin{align*}\text{Length} & = b + 3, \ \text{Width} = b + 2\\ \text{Area} & = (b + 3)(b + 2)\\ & = b^2+2b+3b+6\\ & = b^2 + 5b+6\end{align*}

b) We could add up the areas of the blue and orange rectangles, but it’s easier to just find the area of the whole big rectangle and subtract the area of the yellow rectangle.

\begin{align*}\text{Area of big rectangle} & = 20(12) = 240\\ \text{Area of yellow rectangle} & = (12-x)(20-2x)\\ & = 240-24x-20x+2x^2\\ & = 240-44x+2x^2\\ & = 2x^2-44x+240\end{align*}

The desired area is the difference between the two:

\begin{align*}\text{Area} & = 240-(2x^2-44x+240)\\ & = 240 + (-2x^2+44x-240)\\ & = 240 -2x^2 +44x-240\\ & = -2x^2+44x\end{align*}

c) The volume of this shape = (area of the base)(height).

\begin{align*}\text{Area of the base} & = x(x+2)\\ & = x^2+2x\\ \text{Height} & = 2x + 1\\ \text{Volume} & =(x^2+2x)(2x+1)\\ & = 2x^3+x^2+4x^2+2x\\ & = 2x^3+5x^2+2x\end{align*}

d) The volume of this shape = (area of the base)(height).

\begin{align*}\text{Area of the base} & = (4a-3)(2a+1)\\ & = 8a^2+4a-6a-3\\ & = 8a^2-2a-3\\ \text{Height} & = a + 4\\ \text{Volume} & = (8a^2-2a-3)(a+4)\end{align*}

Let’s multiply using the vertical method:

\begin{align*}& \qquad \quad \ \ \ 8a^2-2a-3\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; a+4\;\;}\\ & \qquad \quad \ 32a^2-8a-12\\ & \underline{8a^3\;-\;\;2a^2\;-3a \;\;\;\;\;\;\;\;\;\;}\\ & 8a^3+\ \ 30a^2-11a-12\end{align*}

The volume is \begin{align*}8a^3+30a^2-11a-12\end{align*}.

Review Questions

Multiply the following monomials.

  1. \begin{align*}(2x)(-7x)\end{align*}
  2. \begin{align*}(10x)(3xy)\end{align*}
  3. \begin{align*}(4mn)(0.5nm^2)\end{align*}
  4. \begin{align*}(-5a^2b)(-12a^3b^3)\end{align*}
  5. \begin{align*}(3xy^2z^2)(15x^2yz^3)\end{align*}

Multiply and simplify.

  1. \begin{align*}17(8x-10)\end{align*}
  2. \begin{align*}2x(4x-5)\end{align*}
  3. \begin{align*}9x^3(3x^2-2x+7)\end{align*}
  4. \begin{align*}3x(2y^2+y-5)\end{align*}
  5. \begin{align*}10q(3q^2r+5r)\end{align*}
  6. \begin{align*}-3a^2b(9a^2-4b^2)\end{align*}
  7. \begin{align*}(x-3)(x+2)\end{align*}
  8. \begin{align*}(a+b)(a-5)\end{align*}
  9. \begin{align*}(x+2)(x^2-3)\end{align*}
  10. \begin{align*}(a^2+2)(3a^2-4)\end{align*}
  11. \begin{align*}(7x-2)(9x-5)\end{align*}
  12. \begin{align*}(2x-1)(2x^2-x+3)\end{align*}
  13. \begin{align*}(3x+2)(9x^2-6x+4)\end{align*}
  14. \begin{align*}(a^2+2a-3)(a^2-3a+4)\end{align*}
  15. \begin{align*}3(x-5)(2x+7)\end{align*}
  16. \begin{align*}5x(x+4)(2x-3)\end{align*}

Find the areas of the following figures.

Find the volumes of the following figures.

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