body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

A1

Algebra 1 View details

4. Special Products of Polynomials

Continue to next lesson

Chapter 7

4.

Special Products of Polynomials

Understanding special products in algebra, particularly those of binomials, is pivotal for various mathematical applications. These patterns simplify complex calculations, especially in real-world scenarios like determining a garden's area. The lesson sheds light on how the degree and leading coefficient of polynomial multiplication are determined. For instance, when dealing with binomials, certain patterns emerge, making calculations more intuitive. Such knowledge is invaluable for students and professionals alike, aiding in tasks ranging from architectural designs to advanced mathematical research.

Lesson Settings & Tools

	10 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Special Products of Polynomials

Slide of 10

The product of some specific binomials can follow certain patterns. These patterns can make calculations easier in contextual situations, such as when calculating a garden's area. This lesson will discuss some of these patterns and how the degree and leading coefficient of the multiplication of polynomials are determined.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Seeing Patterns in the Product of Binomials

Calculate the product of the following binomials.

special products

Is there a formula to calculate the product of the form

(a + b) (a + b) ?

What about the product of the form

(a - b) (a - b) ?

Or the product of the form

(a + b) (a - b) ?

Discussion

The Square of a Binomial

When a binomial is squared, the resulting expression is a perfect square trinomial.

$(a + b)^{2} = a^{2} + 2 a b + b^{2} (a - b)^{2} = a^{2} - 2 a b + b^{2}$

For simplicity, depending on the sign of the binomial, these two identities can be expressed as one.

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

Proof

This rule will be first proven for $(a + b)^{2}$ and then for $(a - b)^{2} .$

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

This identity can be shown by first rewriting the square as a product.

(a + b)^{2}

PowToProdTwoFac

$a^{2} = a \cdot a$

(a + b) (a + b)

▼

Multiply parentheses

Distribute $(a + b)$

a (a + b) + b (a + b)

Distribute $a$

a^{2} + a b + b (a + b)

Distribute $b$

a^{2} + a b + b a + b^{2}

CommutativePropMult

Commutative Property of Multiplication

a^{2} + a b + a b + b^{2}

Add terms

a^{2} + 2 a b + b^{2}

It has been shown that

(a + b)^{2} = a^{2} + 2 a b + b^{2} .

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

In this case, when one term of the binomial is subtracted from the other, the middle term of the perfect square trinomial will instead be negative.

(a + b)^{2}

PowToProdTwoFac

$a^{2} = a \cdot a$

(a - b) (a - b)

▼

Multiply parentheses

Distribute $(a - b)$

a (a - b) - b (a - b)

Distribute $a$

a^{2} - a b - b (a - b)

Distribute $- b$

a^{2} - a b - b a + b^{2}

CommutativePropMult

Commutative Property of Multiplication

a^{2} - a b - a b + b^{2}

Subtract terms

a^{2} - 2 a b + b^{2}

It has been shown that

(a - b)^{2} = a^{2} - 2 a b + b^{2} .

Example

Using the Square of a Binomial to Represent Areas

Izabella wants to change the decorations in her room. She has two square posters of equal size on her wall that she is thinking of changing. She wants to replace one with a poster that is

2

feet longer on each side than the current poster. The second poster will be replaced by one that is

1

foot smaller on each side.

posters

Izabella wants to find an expression in terms of the variable

x

for the difference between the areas. Help her find this expression. Write the answer as a polynomial in standard form.

Hint

Calculate the area of a square with side lengths $x + 2$ feet. Then, calculate the area of a square with side lengths $x - 1$ feet. Finally, find the difference between these areas.

Solution

The area of a square is obtained by squaring its side length.

A = s^{2}

Consider a square of side length

x

feet. If the side lengths are increased by

2

feet, then the length of the new sides is

x + 2

feet.

Therefore, the area of this new square is calculated by squaring

x + 2 .

To do this, the formula for the square of a binomial can be used.

A = (x + 2)^{2}

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

A = x^{2} + 2 x (2) + 2^{2}

▼

Simplify right-hand side

CommutativePropMult

Commutative Property of Multiplication

A = x^{2} + 2 (2) x + 2^{2}

Calculate power

A = x^{2} + 2 (2) (x) + 4

Multiply

A = x^{2} + 4 x + 4

Similarly, if the side lengths are decreased by

1

foot, then the length of the new sides is

x - 1

feet.

The area of this new square is calculated by squaring

x - 1 .

Again, the formula for the square of a binomial can be used.

A = (x - 1)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

A = x^{2} - 2 x (1) + 1^{2}

▼

Simplify right-hand side

Identity Property of Multiplication

A = x^{2} - 2 x + 1^{2}

$1^{a} = 1$

A = x^{2} - 2 x + 1

Finally, to find the difference of the areas in terms of

x,

the expression

x^{2} - 2 x + 1

will be subtracted from

x^{2} + 4 x + 4 .

x^{2} + 4 x + 4 - (x^{2} - 2 x + 1)

Distribute $- 1$

x^{2} + 4 x + 4 - x^{2} + 2 x - 1

Add and subtract terms

6 x + 3

The difference of the areas, in terms of

x,

is

6 x + 3

square feet.

Example

Representing the Area of Praça do Comércio as the Square of a Binomial

The Praça do Comércio is the astonishing main square of Lisbon, the gorgeous capital city of Portugal. Facing the Tagus River, this main court is in the shape of a square with a side length of

3 x + y^{2}

meters.

lisbon square

External credits: Deensel

While researching the city, Izabella wonders about the area of the Praça do Comércio. Find an expression in terms of the variables

x

and

y

for the area of the main square. Write the answer as a polynomial in standard form.

Hint

The area of a square is calculated by squaring its side length.

Solution

The area of a square is calculated by squaring its side length.

A = s^{2}

Therefore, to find the area of the Praça do Comércio, its side length

3 x + y^{2}

must be squared. Since the expression to be squared is a binomial, the formula for the square of a binomial can be used.

(3 x + y^{2})^{2}

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

(3 x)^{2} + 2 (3 x) y^{2} + (y^{2})^{2}

$(a b)^{m} = a^{m} b^{m}$

9 x^{2} + 2 (3 x) y^{2} + (y^{2})^{2}

$(a^{m})^{n} = a^{m \cdot n}$

9 x^{2} + 2 (3 x) y^{2} + y^{4}

Multiply

9 x^{2} + 6 x y^{2} + y^{4}

The area of the square is

9 x^{2} + 6 x y^{2} + y^{4}

square meters.

Discussion

Conjugate Binomials

If two binomials differ only in the sign of one of their terms, they are called conjugate binomials.

The binomials $a + b$ and $a - b$ are conjugate binomials.

Here are some examples.

Conjugate Binomials x + 1 3 x + y x^{2} + 2 y 2 x y + 10 3 x^{2} y + y^{3} x and and and and and x - 1 3 x - y x^{2} - 2 y 2 x y - 10 3 x^{2} y - y^{3} x

Rule

Product of a Conjugate Pair of Binomials

The product of two conjugate binomials is the difference of two squares.

$(a + b) (a - b) = a^{2} - b^{2}$

Proof

This identity can be proved by using the Distributive Property to multiply the binomials.

(a + b) (a - b)

Distribute $(a - b)$

a (a - b) + b (a - b)

Distribute $a$

a^{2} - a b + b (a - b)

Distribute $b$

a^{2} - a b + b a - b^{2}

CommutativePropMult

Commutative Property of Multiplication

a^{2} - a b + a b - b^{2}

Add terms

a^{2} - b^{2}

Therefore, the product of a binomial and its conjugate is the difference of two squares.

Example

Area of a Poster

After researching Praça do Comércio, Izabella decided to buy a poster of it to hang in her room. She is deciding between two posters.

vegetable garden

External credits: Rehman Abubakr, João Eduardo

One poster Izabella is interested in is a square of side length

x

feet. The other is a rectangle of length

x + 2

and width

x - 2

feet. Determine which shape has a greater area.

Calculate the difference between the areas.

Hint

The area of a square is calculated by squaring its side length. The area of a rectangle is calculated by multiplying the length by the width.

Solution

To determine which shape has a greater area, both areas will be calculated. The area of a square is calculated by squaring its side length. For the given square, the side length is

x

feet.

Area of the Square x^{2} ft^{2}

The area of a rectangle is calculated by multiplying the length by the width. For the given rectangle, the length and the width are

x + 2

and

x - 2

feet, respectively. These two expressions are conjugate binomials, so the formula for the product of a conjugate pair of binomials can be used.

(x + 2) (x - 2)

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

x^{2} - 2^{2}

Calculate power

x^{2} - 4

The area of the rectangle has been found.

Area of the Rectangle x^{2} - 4 ft^{2}

Since

x^{2}

is greater than

x^{2} - 4,

the area of the square is greater than the area of the rectangle. Finally, to find the difference,

x^{2} - 4

will be subtracted from

x^{2} .

x^{2} - (x^{2} - 4)

Distribute $- 1$

x^{2} - x^{2} + 4

Subtract term

4

The difference between the areas is

4

square feet.

Example

Leading Coefficient and Degree of a Special Product

When multiplying or squaring binomials, the degree and the leading coefficient of the resulting polynomial may be of interest.

formulas

Calculate the degree and the leading coefficient of each resulting polynomial.

a

(3 x^{3} + x^{2})^{2}

b

(x^{5} - 5 x)^{2}

c

(2 x^{3} + 1) (2 x^{3} - 1)

Hint

a Use the formula for the square of a binomial.

b Use the formula for the square of a binomial.

c Use the formula for multiplying conjugate binomials.

Solution

a To find the degree and the leading coefficient of the resulting polynomial, the formula for squaring a binomial will be used.

(a + b)^{2} = a^{2} + 2 a b + b^{2}

In the given binomial,

a = 3 x^{3}

and

b = x^{2} .

(3 x^{3} + x^{2})^{2}

▼

Simplify

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

(3 x^{3})^{2} + 2 (3 x^{3}) x^{2} + (x^{2})^{2}

$(a b)^{m} = a^{m} b^{m}$

9 (x^{3})^{2} + 2 (3 x^{3}) x^{2} + (x^{2})^{2}

$(a^{m})^{n} = a^{m \cdot n}$

9 x^{6} + 2 (3 x^{3}) x^{2} + x^{4}

Multiply

9 x^{6} + 6 x^{3} x^{2} + x^{4}

$a^{m} \cdot a^{n} = a^{m + n}$

9 x^{6} + 6 x^{5} + x^{4}

The leading coefficient is

9

and the degree is

6 .

b Again, to find the degree and the leading coefficient of the resulting polynomial, the formula for squaring a binomial will be used.

(a - b)^{2} = a^{2} - 2 a b + b^{2}

In the given binomial,

a = x^{5}

and

b = 5 x .

(x^{5} - 5 x)^{2}

▼

Simplify

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

(x^{5})^{2} - 2 x^{5} (5 x) + (5 x)^{2}

$(a^{m})^{n} = a^{m \cdot n}$

x^{10} - 2 x^{5} (5 x) + (5 x)^{2}

CommutativePropMult

Commutative Property of Multiplication

x^{10} - 2 (5 x) x^{5} + (5 x)^{2}

Multiply

x^{10} - 10 x \cdot x^{5} + (5 x)^{2}

$a \cdot a^{m} = a^{1 + m}$

x^{10} - 10 x^{6} + (5 x)^{2}

$(a b)^{m} = a^{m} b^{m}$

x^{10} - 10 x^{6} + 25 x^{2}

Identity Property of Multiplication

1 x^{10} - 10 x^{6} + 25 x^{2}

The leading coefficient is

1

and the degree is

10 .

c In this case, to find the leading coefficient and the degree of the resulting polynomial, two conjugate binomials must be multiplied.

(a + b) (a - b) = a^{2} - b^{2}

Here,

a = 2 x^{3}

and

b = 1 .

(2 x^{3} + 1) (2 x^{3} - 1)

▼

Simplify

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

(2 x^{3})^{2} - 1^{2}

$(a b)^{m} = a^{m} b^{m}$

4 (x^{3})^{2} - 1^{2}

$(a^{m})^{n} = a^{m \cdot n}$

4 x^{6} - 1^{2}

$1^{a} = 1$

4 x^{6} - 1

The leading coefficient is

4

and the degree is

6 .

Pop Quiz

Finding the Leading Coefficient and Degree of Polynomials

State the degree and the leading coefficient of the resulting polynomial after squaring the binomial or multiplying the conjugate binomials.

degree or leading coefficient

Closure

The Cube of a Binomial

The formulas seen in this lesson can be useful to derive other formulas. For example the formula for the square of a binomial can be used to obtain the formula for the cube of a binomial.

(a + b)^{2} = a^{2} + 2 a b + b^{2}

Multiplying this equation by

(a + b)

will give a rule for the cube of a binomial as it creates a rule for

(a + b)^{3} .

(a + b)^{2} = a^{2} + 2 a b + b^{2}

$LHS \cdot (a + b) = RHS \cdot (a + b)$

(a + b)^{2} (a + b) = (a^{2} + 2 a b + b^{2}) (a + b)

CommutativePropMult

Commutative Property of Multiplication

(a + b) (a + b)^{2} = (a^{2} + 2 a b + b^{2}) (a + b)

$a \cdot a^{m} = a^{1 + m}$

(a + b)^{3} = (a^{2} + 2 a b + b^{2}) (a + b)

▼

Simplify right-hand side

Distribute $(a + b)$

(a + b)^{3} = a^{2} (a + b) + 2 a b (a + b) + b^{2} (a + b)

Distribute $a^{2}$

(a + b)^{3} = a^{3} + a^{2} b + 2 a b (a + b) + b^{2} (a + b)

Distribute $2 a b$

(a + b)^{3} = a^{3} + a^{2} b + 2 a^{2} b + 2 a b^{2} + b^{2} (a + b)

Distribute $b^{2}$

(a + b)^{3} = a^{3} + a^{2} b + 2 a^{2} b + 2 a b^{2} + a b^{2} + b^{3}

Add terms

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

To find a rule for

(a - b)^{3},

b

can be replaced with

- b

in the obtained formula.

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

SubstituteExpressions

Substitute expressions

(a + (- b))^{3} = a^{3} + 3 a^{2} (- b) + 3 a (- b)^{2} + (- b)^{3}

▼

Simplify

$a (- b) = - a \cdot b$

(a + (- b))^{3} = a^{3} + (- 3 a^{2} b) + 3 a (- b)^{2} + (- b)^{3}

NegBaseToNegPow

$(- a)^{3} = - a^{3}$

(a + (- b))^{3} = a^{3} + (- 3 a^{2} b) + 3 a (- b)^{2} + (- b^{3})

$a + (- b) = a - b$

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a (- b)^{2} - b^{3}

NegBaseToPosPow

$(- a)^{2} = a^{2}$

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

By using the formula for the square of a binomial, two formulas for the cube of a binomial were derived. For simplicity, depending on the sign of the binomial, these two identities can be expressed as one.

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} (a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} ⇓ (a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}

Special Products of Polynomials

Exercise 3.1

Find the value of

c

that makes the polynomial a perfect square trinomial.

81 x^{2} - 198 x + c

A perfect square trinomial is an expression that is the result of squaring either a sum or a difference of two monomials. ( a± b)^2 &= a^2± 2 a b+ b^2_() & Perfect square trinomial Now, we will consider the given trinomial. 81x^2 - 198x + c Since the sign of the linear term is negative, we are looking for the square of a difference of monomials. To find the value of c, we will first find the values of a and b by using the Properties of Exponents. Let's start by rewriting 81x^2 to find a.

We have found that a= 9x. Now, we will rewrite the middle term 198x to find the value of b. Remember that we are looking for 2 a b. Therefore, we will split 198x into factors in such a way including 9x as a factor in it. 198x ⇒ 2( 9x)( 11) This means that the value of b is 11. Let's substitute the values we know into the formula for the square of a binomial with a negative sign. ( a- b)^2 & = a^2-2 a b+ b^2 [0.5em] & = 9x^2-2( 9x)( 11)+c We can deduce that, in order for the given trinomial to be a perfect square trinomial, the value of c must be 11^2= 121. Since we have completed the given task we can stop here, but let's substitute c=121 into our expression and express the trinomial as the square of a binomial.

Special Products of Polynomials

Recommended exercises

To get personalized recommended exercises, please login first.

Return to lesson.