<meta http-equiv="refresh" content="1; url=/nojavascript/"> Multiplication of Polynomials by Binomials ( Read ) | Algebra | CK-12 Foundation
Dismiss
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Multiplication of Polynomials by Binomials

%
Best Score
Practice Multiplication of Polynomials by Binomials
Practice
Best Score
%
Practice Now
Multiplication of Polynomials by Binomials
 0  0  0

What if you had two polynomials like 3x^3 + 2x^2 and x^2 - 1 ? How could you multiply them? After completing this Concept, you'll be able to use the Distributive Property to multiply one polynomial by another.

Watch This

CK-12 Foundation: 0904S Multiplying a Polynomial by a Polynomial

This Khan Academy video shows how multiplying two binomials together is related to the distributive property.

Khan Academy: Level 1 multiplying expressions

Guidance

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 (a+b)(c+d) .

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 c , multiply it by d , and then add those two products together: (a+b)(c+d)=(a+b) \cdot c+(a+b) \cdot d .

We can rewrite this expression as c(a+b)+d(a+b) . 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 c(a+b)+d(a+b)=ca+cb+da+db.

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 A

Multiply and simplify: (2x+1)(x+3)

Solution

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 parentheses by all terms in the second parentheses and add them to the previous terms.

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

(2x)(x)+(2x)(3)+(1)(x)+(1)(3)=2x^2+6x+x+3=2x^2+7x+3

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 B

Multiply and simplify:

a) (4x-5)(x-20)

b) (3x-2)(3x+2)

c) (3x^2+2x-5)(2x-3)

d) (x^2-9)(4x^4+5x^2-2)

Solution

a) With horizontal multiplication this would be

(4x-5)(x-20)&=(4x)(x)+(4x)(-20)+(-5)(x)+(-5)(-20)\\ &=4x^2-80x-5x+100\\ &=4x^2-85x+100

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

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

Both techniques result in the same answer: 4x^2-85x+100 . We’ll use vertical multiplication for the other problems.

b)

& \qquad \ \ 3x-2\\& \underline{\;\;\;\;\;\;\;\;\; 3x+2}\\& \qquad \ \ 6x-4\\& \underline{9x^2-6x \;\;\;\;\;\;}\\& 9x^2+0x-4

The answer is 9x^2-4 .

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

& \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

The answer is 6x^3-5x^2-16x+15 .

d) Set up the multiplication vertically and leave gaps for missing powers of x :

& \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

The answer is 4x^6-31x^4-47x^2+18 .

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 C

a) Find the areas of the figure:

b) Find the volumes of the figure:

Solutions:

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

For the big rectangle:

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

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

\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

Watch this video for help with the Examples above.

CK-12 Foundation: Multiplying a Polynomial by a Polynomial

Vocabulary

  • A binomial is a polynomial with two terms.
  • The Distributive Property for Binomials: The Distributive Property says that the term in front of the parentheses multiplies with each term inside the parentheses separately. Then, we add the results of the products.

(a+b)(c+d)&=c\cdot (a+b)+d\cdot (a+b)\\ &=c\cdot a+c \cdot b+d \cdot a+d \cdot b \\ &= \ ca+cb+da+db

Guided Practice

1. Find the areas of the figure:

2. Find the volumes of the figure:

Solutions:

1. 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.

\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

The desired area is the difference between the two:

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

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

\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)

Let’s multiply using the vertical method:

& \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

The volume is 8a^3+30a^2-11a-12 .

Practice

Multiply and simplify.

  1. (x-3)(x+2)
  2. (a+b)(a-5)
  3. (x+2)(x^2-3)
  4. (a^2+2)(3a^2-4)
  5. (7x-2)(9x-5)
  6. (2x-1)(2x^2-x+3)
  7. (3x+2)(9x^2-6x+4)
  8. (a^2+2a-3)(a^2-3a+4)
  9. 3(x-5)(2x+7)
  10. 5x(x+4)(2x-3)

Find the areas of the following figures.

Find the volumes of the following figures.

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...
ShareThis Copy and Paste

Original text