{{ 'ml-label-loading-course' | message }}

{{ tocSubheader }}

{{ 'ml-toc-proceed-mlc' | message }}

{{ 'ml-toc-proceed-tbs' | message }}

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.intro.summary }}

Show less Show more Lesson Settings & Tools

| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |

| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |

| {{ 'ml-lesson-time-estimation' | message }} |

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

Discussion

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

$(a+b)_{2}=a_{2}+2ab+b_{2}(a−b)_{2}=a_{2}−2ab+b_{2}$

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

$(a±b)_{2}=a_{2}±2ab+b_{2}$

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

$(a+b)_{2}$

PowToProdTwoFac

$a_{2}=a⋅a$

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

▼

Multiply parentheses

Distr

Distribute $(a+b)$

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

Distr

Distribute $a$

$a_{2}+ab+b(a+b)$

Distr

Distribute $b$

$a_{2}+ab+ba+b_{2}$

CommutativePropMult

Commutative Property of Multiplication

$a_{2}+ab+ab+b_{2}$

AddTerms

Add terms

$a_{2}+2ab+b_{2}$

$(a+b)_{2}$

PowToProdTwoFac

$a_{2}=a⋅a$

$(a−b)(a−b)$

▼

Multiply parentheses

Distr

Distribute $(a−b)$

$a(a−b)−b(a−b)$

Distr

Distribute $a$

$a_{2}−ab−b(a−b)$

Distr

Distribute $-b$

$a_{2}−ab−ba+b_{2}$

CommutativePropMult

Commutative Property of Multiplication

$a_{2}−ab−ab+b_{2}$

SubTerms

Subtract terms

$a_{2}−2ab+b_{2}$

Example

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. ### Hint

### Solution

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Difference in areas:","formTextAfter":"square feet","answer":{"text":["6x+3","3(2x+1)"]}}

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.

The area of a square is obtained by squaring its side length.
Similarly, if the side lengths are decreased by $1$ foot, then the length of the new sides is $x−1$ feet.
Finally, to find the difference of the areas in terms of $x,$ the expression $x_{2}−2x+1$ will be subtracted from $x_{2}+4x+4.$
The difference of the areas, in terms of $x,$ is $6x+3$ square feet.

$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

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$A=x_{2}+2x(2)+2_{2}$

▼

Simplify right-hand side

CommutativePropMult

Commutative Property of Multiplication

$A=x_{2}+2(2)x+2_{2}$

CalcPow

Calculate power

$A=x_{2}+2(2)(x)+4$

Multiply

Multiply

$A=x_{2}+4x+4$

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

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$A=x_{2}−2x(1)+1_{2}$

▼

Simplify right-hand side

$A=x_{2}−2x+1$

$x_{2}+4x+4−(x_{2}−2x+1)$

Distr

Distribute $-1$

$x_{2}+4x+4−x_{2}+2x−1$

AddSubTerms

Add and subtract terms

$6x+3$

Example

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 $3x+y_{2}$ meters.
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

### Solution

External credits: Deensel

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Area:","formTextAfter":"square meters","answer":{"text":["9x^2+6xy^2+y^4","9x^2+6y^2x+y^4","y^4+6xy^2+9x^2","y^4+6y^2x+9x^2"]}}

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

The area of a square is calculated by squaring its side length.
The area of the square is $9x_{2}+6xy_{2}+y_{4}$ square meters.

$A=s_{2} $

Therefore, to find the area of the Praça do Comércio, its side length $3x+y_{2}$ must be squared. Since the expression to be squared is a binomial, the formula for the square of a binomial can be used.
$(3x+y_{2})_{2}$

ExpandPosPerfectSquare

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$(3x)_{2}+2(3x)y_{2}+(y_{2})_{2}$

PowProdII

$(ab)_{m}=a_{m}b_{m}$

$9x_{2}+2(3x)y_{2}+(y_{2})_{2}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$9x_{2}+2(3x)y_{2}+y_{4}$

Multiply

Multiply

$9x_{2}+6xy_{2}+y_{4}$

Discussion

Rule

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

$(a+b)(a−b)=a_{2}−b_{2}$

This identity can be proved by using the Distributive Property to multiply the binomials.
Therefore, the product of a binomial and its conjugate is the difference of two squares.

$(a+b)(a−b)$

Distr

Distribute $(a−b)$

$a(a−b)+b(a−b)$

Distr

Distribute $a$

$a_{2}−ab+b(a−b)$

Distr

Distribute $b$

$a_{2}−ab+ba−b_{2}$

CommutativePropMult

Commutative Property of Multiplication

$a_{2}−ab+ab−b_{2}$

AddTerms

Add terms

$a_{2}−b_{2}$

Example

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

### Solution

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.

{"type":"choice","form":{"alts":["Square","Rectangle"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Difference between the areas:","formTextAfter":"square feet","answer":{"text":["4"]}}

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.

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 Squarex_{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.
The area of the rectangle has been found.
$Area of the Rectanglex_{2}−4ft_{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 $x_{2}.$
The difference between the areas is $4$ square feet.
Example

When multiplying or squaring binomials, the degree and the leading coefficient of the resulting polynomial may be of interest.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Degree:","formTextAfter":null,"answer":{"text":["6"]}} ### Hint

### Solution

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

a $(3x_{3}+x_{2})_{2}$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Degree:","formTextAfter":null,"answer":{"text":["6"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Leading Coefficient:","formTextAfter":null,"answer":{"text":["9"]}}

b $(x_{5}−5x)_{2}$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Degree:","formTextAfter":null,"answer":{"text":["10"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Leading Coefficient:","formTextAfter":null,"answer":{"text":["1"]}}

c $(2x_{3}+1)(2x_{3}−1)$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Leading Coefficient:","formTextAfter":null,"answer":{"text":["4"]}}

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.

a To find the degree and the leading coefficient of the resulting polynomial, the formula for squaring a binomial will be used.
The leading coefficient is $9$ and the degree is $6.$

$(a+b)_{2}=a_{2}+2ab+b_{2} $

In the given binomial, $a=3x_{3}$ and $b=x_{2}.$
$(3x_{3}+x_{2})_{2}$

▼

Simplify

ExpandPosPerfectSquare

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$(3x_{3})_{2}+2(3x_{3})x_{2}+(x_{2})_{2}$

PowProdII

$(ab)_{m}=a_{m}b_{m}$

$9(x_{3})_{2}+2(3x_{3})x_{2}+(x_{2})_{2}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$9x_{6}+2(3x_{3})x_{2}+x_{4}$

Multiply

Multiply

$9x_{6}+6x_{3}x_{2}+x_{4}$

MultPow

$a_{m}⋅a_{n}=a_{m+n}$

$9x_{6}+6x_{5}+x_{4}$

b Again, to find the degree and the leading coefficient of the resulting polynomial, the formula for squaring a binomial will be used.
The leading coefficient is $1$ and the degree is $10.$

$(a−b)_{2}=a_{2}−2ab+b_{2} $

In the given binomial, $a=x_{5}$ and $b=5x.$
$(x_{5}−5x)_{2}$

▼

Simplify

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$(x_{5})_{2}−2x_{5}(5x)+(5x)_{2}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$x_{10}−2x_{5}(5x)+(5x)_{2}$

CommutativePropMult

Commutative Property of Multiplication

$x_{10}−2(5x)x_{5}+(5x)_{2}$

Multiply

Multiply

$x_{10}−10x⋅x_{5}+(5x)_{2}$

MultBasePow

$a⋅a_{m}=a_{1+m}$

$x_{10}−10x_{6}+(5x)_{2}$

PowProdII

$(ab)_{m}=a_{m}b_{m}$

$x_{10}−10x_{6}+25x_{2}$

IdPropMult

Identity Property of Multiplication

$1x_{10}−10x_{6}+25x_{2}$

c In this case, to find the leading coefficient and the degree of the resulting polynomial, two conjugate binomials must be multiplied.
The leading coefficient is $4$ and the degree is $6.$

$(a+b)(a−b)=a_{2}−b_{2} $

Here, $a=2x_{3}$ and $b=1.$
$(2x_{3}+1)(2x_{3}−1)$

▼

Simplify

ExpandDiffSquares

$(a+b)(a−b)=a_{2}−b_{2}$

$(2x_{3})_{2}−1_{2}$

PowProdII

$(ab)_{m}=a_{m}b_{m}$

$4(x_{3})_{2}−1_{2}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$4x_{6}−1_{2}$

BaseOne

$1_{a}=1$

$4x_{6}−1$

Pop Quiz

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

Closure

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.
*cube of a binomial* as it creates a rule for $(a+b)_{3}.$
To find a rule for $(a−b)_{3},$ $b$ can be replaced with $-b$ in the obtained formula.

$(a+b)_{2}=a_{2}+2ab+b_{2} $

Multiplying this equation by $(a+b)$ will give a rule for the $(a+b)_{2}=a_{2}+2ab+b_{2}$

MultEqn

$LHS⋅(a+b)=RHS⋅(a+b)$

$(a+b)_{2}(a+b)=(a_{2}+2ab+b_{2})(a+b)$

CommutativePropMult

Commutative Property of Multiplication

$(a+b)(a+b)_{2}=(a_{2}+2ab+b_{2})(a+b)$

MultBasePow

$a⋅a_{m}=a_{1+m}$

$(a+b)_{3}=(a_{2}+2ab+b_{2})(a+b)$

▼

Simplify right-hand side

Distr

Distribute $(a+b)$

$(a+b)_{3}=a_{2}(a+b)+2ab(a+b)+b_{2}(a+b)$

Distr

Distribute $a_{2}$

$(a+b)_{3}=a_{3}+a_{2}b+2ab(a+b)+b_{2}(a+b)$

Distr

Distribute $2ab$

$(a+b)_{3}=a_{3}+a_{2}b+2a_{2}b+2ab_{2}+b_{2}(a+b)$

Distr

Distribute $b_{2}$

$(a+b)_{3}=a_{3}+a_{2}b+2a_{2}b+2ab_{2}+ab_{2}+b_{3}$

AddTerms

Add terms

$(a+b)_{3}=a_{3}+3a_{2}b+3ab_{2}+b_{3}$

$(a+b)_{3}=a_{3}+3a_{2}b+3ab_{2}+b_{3}$

SubstituteExpressions

Substitute expressions

$(a+(-b))_{3}=a_{3}+3a_{2}(-b)+3a(-b)_{2}+(-b)_{3}$

▼

Simplify

MultPosNeg

$a(-b)=-a⋅b$

$(a+(-b))_{3}=a_{3}+(-3a_{2}b)+3a(-b)_{2}+(-b)_{3}$

NegBaseToNegPow

$(-a)_{3}=-a_{3}$

$(a+(-b))_{3}=a_{3}+(-3a_{2}b)+3a(-b)_{2}+(-b_{3})$

AddNeg

$a+(-b)=a−b$

$(a−b)_{3}=a_{3}−3a_{2}b+3a(-b)_{2}−b_{3}$

NegBaseToPosPow

$(-a)_{2}=a_{2}$

$(a−b)_{3}=a$