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Multiplication of Monomials by Polynomials

Distribute single terms by multiplying them by other terms

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Multiplication of Monomials by Polynomials

Did you know that the formula for the volume of a pyramid is \begin{align*}V= \frac{1}{3}Bh\end{align*}V=13Bh, where \begin{align*}B\end{align*}B is the area of the base of the pyramid and \begin{align*}h\end{align*}h is the pyramid's height? What would the volume of the pyramid be if the base of a pyramid were \begin{align*}x^2 + 6x + 8\end{align*}x2+6x+8 and the height were \begin{align*}9x\end{align*}9x?

Multiplying Monomials and Polynomials

When multiplying polynomials together, we must remember the exponent rules such as the Product Rule. This rule says that if we multiply expressions that have the same base, we just add the exponents and keep the base unchanged. If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any numbers. We also apply the product rule on each variable separately. For example:

\begin{align*}(2x^2 y^3) \times (3x^2 y)=(2\cdot3) \times (x^2 \cdot x^2) \times (y^3 \cdot y)=6x^4 y^4\end{align*}(2x2y3)×(3x2y)=(23)×(x2x2)×(y3y)=6x4y4

Multiplying a Monomial by a Monomial

Multiplying a monomial by a monomial is the simplest of polynomial multiplications.

Let's multiply the following monomials:

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

\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

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

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

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

\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

Multiplying Monomials by Polynomials

To multiply monomials by polynomials, we use the Distributive Property. Recall that the Distributive Property states that for any expressions \begin{align*}a, \ b\end{align*}a, b, and \begin{align*}c\end{align*}c, \begin{align*}a(b+c)=ab+ac\end{align*}a(b+c)=ab+ac.

This property can be used for numbers as well as variables. This property is best illustrated by an area problem. We can find the area of the big rectangle shown below in two ways.

One way is to use the formula for the area of a rectangle.

\begin{align*}Area \ of \ the \ big \ rectangle & = Length \times Width\\ Length & = a, \ Width = b + c\\ Area & = a \times (b + c)\end{align*}Area of the big rectangleLengthArea=Length×Width=a, Width=b+c=a×(b+c)

The area of the big rectangle can also be found by adding the areas of the two smaller rectangles.

\begin{align*}Area \ of \ red \ rectangle & = ab\\ Area \ of \ blue \ rectangle & = ac\\ Area \ of \ big \ rectangle & = ab + ac\end{align*}Area of red rectangleArea of blue rectangleArea of big rectangle=ab=ac=ab+ac

This means that \begin{align*}a(b+c)=ab+ac\end{align*}a(b+c)=ab+ac.

In general, if we have a number or variable in front of a parenthesis, this means that each term in the parenthesis is multiplied by the expression in front of the parenthesis.

\begin{align*}a(b+c+d+e+f+\ldots)=ab+ac+ad+ae+af+ \ldots\end{align*}a(b+c+d+e+f+)=ab+ac+ad+ae+af+ The “...” means “and so on.”

Let's multiply \begin{align*}2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\end{align*}2x3y(3x4y2+2x3y10x2+7x+9):

\begin{align*}& 2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\\ & = (2x^3 y)(-3x^4 y^2 )+(2x^3 y)(2x^3 y)+(2x^3 y)(-10x^2 )+(2x^3 y)(7x)+(2x^3 y)(9)\\ & = -6x^7 y^3+4x^6 y^2-20x^5 y+14x^4 y+18x^3 y\end{align*}2x3y(3x4y2+2x3y10x2+7x+9)=(2x3y)(3x4y2)+(2x3y)(2x3y)+(2x3y)(10x2)+(2x3y)(7x)+(2x3y)(9)=6x7y3+4x6y220x5y+14x4y+18x3y

   

 

 

Examples

Example 1

Earlier, you were told that the formula for the volume of a pyramid is \begin{align*}V= \frac{1}{3}Bh\end{align*}, where \begin{align*}B\end{align*} is the area of the base of the pyramid and \begin{align*}h\end{align*} is the pyramid's height. What would the volume of the pyramid be if the area of the base were \begin{align*}x^2 + 6x + 8\end{align*} and the height were \begin{align*}9x\end{align*}?

To find the volume of the pyramid with the given area of the base and height, we need to plug in the expressions to the volume equation:

\begin{align*}V &= \frac{1}{3}(x^2 + 6x + 8)(9x)\end{align*}

Now, use the Distributive Property to simplify:

\begin{align*}V &= \frac{1}{3}(x^2 + 6x + 8)(9x)\\ &=\frac{1}{3}(9x)(x^2+6x+8) &\text{rearrange the terms using the Associative Property}\\ &=3x(x^2+6x+8) &\text{Multiply the first two terms}\\ &=3x^3 + 18x^2 + 24x&\text{Using the Distributive Property}\end{align*} 

The volume of the pyramid with base area \begin{align*}x^2 + 6x + 8\end{align*} and height \begin{align*}9x\end{align*} is \begin{align*}3x^3+18x^2+24x\end{align*}.

Example 2

Multiply \begin{align*}-2a^2b^4(3ab^2+7a^3b-9a+3)\end{align*}.

Multiply the monomial by each term inside the parenthesis:

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

Review

Multiply the following monomials.

  1. \begin{align*}(2x)(-7x)\end{align*}
  2. \begin{align*}4(-6a)\end{align*}
  3. \begin{align*}(-5a^2b)(-12a^3b^3)\end{align*}
  4. \begin{align*}(-5x)(5y)\end{align*}
  5. \begin{align*}y(xy^4)\end{align*}
  6. \begin{align*}(3xy^2z^2)(15x^2yz^3)\end{align*}

Multiply and simplify.

  1. \begin{align*}x^8 (xy^3+3x)\end{align*}
  2. \begin{align*}2x(4x-5)\end{align*}
  3. \begin{align*}6ab(-10a^2 b^3+c^5)\end{align*}
  4. \begin{align*}9x^3(3x^2-2x+7)\end{align*}
  5. \begin{align*}-3a^2b(9a^2-4b^2)\end{align*}

Mixed Review

  1. Give an example of a fourth degree trinomial in the variable \begin{align*}n\end{align*}.
  2. Find the next four terms of the sequence \begin{align*}1,\frac{3}{2},\frac{9}{4},\frac{28}{8}, \ldots\end{align*}
  3. Reece reads three books per week.
    1. Make a table of values for weeks zero through six.
    2. Fit a model to this data.
    3. When will Reece have read 63 books?
  1. Write 0.062% as a decimal.
  2. Evaluate \begin{align*}ab\left ( a+\frac{b}{4} \right )\end{align*} when \begin{align*}a=4\end{align*} and \begin{align*}b=-3\end{align*}.
  3. Solve for \begin{align*}s\end{align*}: \begin{align*}3s(3+6s)+6(5+3s)=21s\end{align*}.

Review (Answers)

To see the Review answers, open this PDF file and look for section 9.3. 

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