Level 2 - Proving Polynomial Identities

Equations that are true for every possible value of the variable are called identities. Some special polynomial equations are identities. Using polynomial identities can be useful when rewriting polynomial expressions. To prove an equation is an identity, it is enough to show that both sides can be written in the same way.

Rule

Square of a Binomial

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

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

(a±b)2=a2±2ab+b2

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

a2=a⋅a

(a+b)(a+b)

Multiply parentheses

Distr

Distribute (a+b)

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

Distr

Distribute a

a2+ab+b(a+b)

Distr

Distribute b

a2+ab+ba+b2

CommutativePropMult

Commutative Property of Multiplication

a2+ab+ab+b2

AddTerms

Add terms

a2+2ab+b2

It has been shown that (a+b)2=a2+2ab+b2.

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

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

a2=a⋅a

(a−b)(a−b)

Multiply parentheses

Distr

Distribute (a−b)

a(a−b)−b(a−b)

Distr

Distribute a

a2−ab−b(a−b)

Distr

Distribute -b

a2−ab−ba+b2

CommutativePropMult

Commutative Property of Multiplication

a2−ab−ab+b2

SubTerms

Subtract terms

a2−2ab+b2

It has been shown that (a−b)2=a2−2ab+b2.

Rewrite the expressions as perfect square trinomials

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Rewrite (x+3)2 and (7−x)2 as perfect square trinomials.

Show Solution

When rewriting (x+3)2 as a perfect square trinomial, keep in mind that it is the square of a binomial.

(x+3)2

ExpandPosPerfectSquare

(a+b)2=a2+2ab+b2

x2+2⋅x⋅3+32

Multiply

x2+6x+32

CalcPow

Calculate power

x2+6x+9

The second expression has a minus sign between the terms – take this into account when rewriting the binomial.

(7−x)2

ExpandNegPerfectSquare

(a−b)2=a2−2ab+b2

72−2⋅7⋅x+x2

CalcPow

Calculate power

49−2⋅7⋅x+x2

Multiply

49−14x+x2

The expressions rewritten as perfect square trinomials are x2+6x+9 and 49−14x+x2, respectively.

Rule

Product of a Conjugate Pair of Binomials

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

(a+b)(a−b)=a2−b2

Proof

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

(a+b)(a−b)

Distr

Distribute (a−b)

a(a−b)+b(a−b)

Distr

Distribute a

a2−ab+b(a−b)

Distr

Distribute b

a2−ab+ba−b2

CommutativePropMult

Commutative Property of Multiplication

a2−ab+ab−b2

AddTerms

Add terms

a2−b2

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

Rule

Sum of Two Cubes

Rule

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

To show the identity

it is enough to rewrite the right-hand side and show that it equals the sum on the left-hand side.

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

MultPar

Multiply parentheses

a⋅a2−a⋅ab+a⋅b2+b⋅a2−b⋅ab+b⋅b2

Multiply

a3−a2b+ab2+a2b−ab2+b3

CommutativePropAdd

Commutative Property of Addition

a3−a2b+a2b+ab2−ab2+b3

SimpTerms

Simplify terms

a3+b3

Rule

Difference of Two Cubes

Rule

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

Showing this identity is easiest done by rewriting the right-hand side.

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

MultPar

Multiply parentheses

a⋅a2+a⋅ab+a⋅b2−b⋅a2−b⋅ab−b⋅b2

Multiply

a3+a2b+ab2−a2b−ab2−b3

CommutativePropAdd

Commutative Property of Addition

a3+a2b−a2b+ab2−ab2−b3

SimpTerms

Simplify terms

a3−b3

Thus, the identity

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

is true.

Solve the equation

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Find all complex solutions to the equation x3−43=0 by using the difference of two cubes.

Show Solution

To solve the equation x3−43=0, notice that the expression on the left-hand side is the difference of two cubes. We can use that to rewrite it as a product.

The Zero Product Property tells us that we can solve the equation by setting each of the factors equal to 0 and solving the resulting equations separately.