Topics
2. Polynomial Expressions and Equations
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United States of America
Grades 9-12

2.03 Polynomial identities

Lesson
Practice

Introduction

Some of the special products we discovered in lesson  1.08 Factoring using appropriate methods  are actually mathematical identities. We will explore what these are in this lesson and how they are different from equations, and we will learn how to prove or verify that they hold true for all values of the variables within them.

Polynomial identities

Mathematical identity

Types of equations that remain true for any values represented by the variables

While an equation might be true for some specific value of variables within it, a mathematical identity is true for every value of the variables within it. Many of the patterns used in factoring polynomials are mathematical identities.

Exploration

A square with side lengths of m. A smaller square with length of n and an area of n squared is cut out from the bottom right corner. The remaining lower smaller rectangle of width n from the bottom is rotated and placed on the left hand side of the square. Speak to your teacher for more details.

In the diagram, a square of side n has been cut out of a square of size m. What remains can be divided into two rectangles, and the lower smaller rectangle can be rotated and re-attached to the left hand side.

  1. What familiar identity can we prove using this diagram?

One mathematical identity we have worked with already is the difference of squares, m^2-n^2=(m-n)(m+n). It is easily proven using a geometric diagram such as the one above.

Another familiar identity we have previously used is {a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2}} when factoring perfect square trinomials. Two more important identities are the sum of cubes and difference of cubes.

Sum of cubes identity

Two perfect cube expressions being added to each other

a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)

Difference of cubes identity

Two perfect cube expressions being subtracted from each other

a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)

In the proof of an identity, it is our job to prove both sides are equal, so we must work with one side of the equation and show that algebraic manipulation leads to the other side. We cannot manipulate both sides of the equation at once because changing both sides assumes that both sides are already equal.

We can extend our knowledge of complex numbers and polynomial identities to find complex factors of polynomials. Using the fact that i^2=-1, we can often rewrite a constant term and factor the expression further. For example,

\begin{aligned}x^2+3&=x^2-3i^2\\&=\left(x+\sqrt{3}i\right)\left(x-\sqrt{3}i\right)\end{aligned}

Examples

Example 1

The perfect square trinomial identity is \left(a+b\right)^2=a^2+2ab+b^2.

a

Prove the identity.

Worked Solution
Create a strategy

We can prove the identity using a geometric diagram.

Apply the idea

We will begin by drawing a square with side lengths of a+b. The area of this square is the left side of the equation, \left(a+b\right)^2.

A square divided into 2 rows of 2 rectangles. The outside of the square is labeled a above the left column and b above the right column. The side is labeled a next to the top row and b next to the bottom row.

Next, we can find the areas of the squares and rectangles within the larger square.

A square divided into 2 rows of 2 rectangles. From top to bottom, the left column of squares is labeled a squared and a b, and the right column of squares is labeled a b and b squaed. The outside of the square is labeled a above the left column and b above the right column. The side is labeled a next to the top row and b next to the bottom row.

The area of the larger square is equal to the sum of the areas of the squares and rectangles within it. Therefore, \left(a+b\right)^2=a^2+2ab+b^2.

Reflect and check

Alternatively, we could prove the identity through algebraic manipulation. We are trying to prove \left(a+b\right)^2=a^2+2ab+b^2 We can only manipulate one side of the equation, so we will expand the left hand side using the distributive property.

\displaystyle \left(a+b\right)^2\displaystyle =\displaystyle \left(a+b\right)\left(a+b\right)Expand the power
\displaystyle =\displaystyle a^2+ab+ba+b^2Distributive property
\displaystyle =\displaystyle a^2+ab+ab+b^2Commutative property of multiplication
\displaystyle =\displaystyle a^2+2ab+b^2Add like terms
b

Use the identity to evaluate 98^2.

Worked Solution
Create a strategy

Because 98 is squared, we can represent it with the left side of the equation.\left(a+b\right)^2=98^2

Apply the idea

When choosing values for a and b, we want to choose values that would be easy to calculate with. One option is to use a=100 and b=-2. \left(100+-2\right)^2=98^2

Now, we can use the right side of the identity to easily calculate the value of 98^2.

\displaystyle \left(a+b\right)^2\displaystyle =\displaystyle a^2+2ab+b^2State the identity
\displaystyle \left(100+-2\right)^2\displaystyle =\displaystyle \left(100\right)^2+2\left(100\right)\left(-2\right)+\left(-2\right)^2Substitute a=100 and b=-2
\displaystyle 98^2\displaystyle =\displaystyle 10\,000-400+4Evaluate the multiplication and powers
\displaystyle 98^2\displaystyle =\displaystyle 9\,604Evaluate the addition and subtraction
Reflect and check

You can use your calculator to confirm this answer, but this is a helpful strategy to use when calculator use is not permitted.

Example 2

Prove x^{n}-1=\left(x-1\right)\left(x^{n-1}+x^{n-2}+\ldots+x+1\right) where n is any positive integer.

Worked Solution
Create a strategy

We can prove this by expanding the right side of the equation.

Apply the idea

We can organize our work nicely by distributing the x first and writing the terms in one line, then distributing the -1 and writing the terms on a second line.

\begin{aligned}(x&-1)\left(x^{n-1}+x^{n-2}+\ldots+x+1\right)\\=x^n&+x^{n-1}+x^{n-2}+\ldots+x^2+x\\&-x^{n-1}-x^{n-2}-\ldots-x^2-x-1\\ \hline =x^{n}&-1\end{aligned}

By organizing our work this way, it was easy to see that many of the terms would add to 0. Therefore, we have proven the identity because the expressions on both sides of the equation are identical for all values of x and any positive integer n.

Reflect and check

This identity will be very useful when working with Geometric Sequences in lesson  5.06 Geometric series  .

Example 3

Write each polynomial as a product of linear factors. Leave your answer in terms of i.

a

4x^4+35x^2-9

Worked Solution
Create a strategy

There is no common factor, so we need to use the method of factoring trinomials when the leading coefficient is not 1.

Recall that we did this in Math 2 using the following steps:

  1. Find the product of the leading coefficient and the constant, ac
  2. Find two numbers that multiply to ac and add to the middle coefficient, b
  3. Factor by grouping
Apply the idea

In this problem, a=4, b=35, c=-9. ac=4\cdot -9=-36\\b=35

Two numbers that multiply to ac and add to b are -1 and 36.

\displaystyle 4x^4+35x^2+9\displaystyle =\displaystyle 4x^4-x^2+36x^2-9Rewrite the middle term
\displaystyle =\displaystyle x^2\left(4x^2-1\right)+9\left(4x^2-1\right)Factor by grouping
\displaystyle =\displaystyle \left(4x^2-1\right)\left(x^2+9\right)Factor out the GCF
\displaystyle =\displaystyle \left(2x+1\right)\left(2x-1\right)\left(x^2+9\right)Difference of squares identity

We need to factor the expression until all factors are linear, but x^2+9 is quadratic. We can use our knowledge of complex numbers to rewrite x^2+9 as a difference of squares using the fact that i^2=-1. In other words, x^2+9=x^2-9i^2. This will allow us to factor the expression further.

\displaystyle \left(2x+1\right)\left(2x-1\right)\left(x^2+9\right)\displaystyle =\displaystyle \left(2x+1\right)\left(2x-1\right)\left(x^2-9i^2\right)Rewrite as difference of squares
\displaystyle =\displaystyle \left(2x+1\right)\left(2x-1\right)\left(x+3i\right)\left(x-3i\right)Difference of squares identity

The full factorization is {4x^4+25x^2+9=\left(2x+1\right)\left(2x-1\right)\left(x+3i\right)\left(x-3i\right)} where all factors are linear.

Reflect and check

Before learning of complex numbers, we would have left our answer as \left(2x+1\right)\left(2x-1\right)\left(x^2+9\right). Complex numbers allows us to factor further which will help us find both the real and complex roots of a polynomial when we solve an equation.

b

4x^2-12ix-9

Worked Solution
Create a strategy

The middle term has a coefficient of -12i. If we rewrite the last term using i^2=-1, we might be able to factor the expression.

Apply the idea

4x^2-12ix-9=4x^2-12ix+9i^2

Now, we can factor this using the perfect square trinomial identity \left(a-b\right)^2=a^2-2ab+b^2 where a=2x and b=3i.

4x^2-12ix+9i^2=\left(2x-3i\right)^2

Therefore, 4x^2-12ix-9=\left(2x-3i\right)^2.

Reflect and check

We will verify our answer by expanding \left(2x-3i\right)^2:

\displaystyle \left(2x-3i\right)^2\displaystyle =\displaystyle \left(2x-3i\right)\left(2x-3i\right)Expand the power
\displaystyle =\displaystyle 4x^2-6ix-6ix+9i^2Distributive property
\displaystyle =\displaystyle 4x^2-12ix+9i^2Combine like terms
\displaystyle =\displaystyle 4x^2-12ix+9\left(-1\right)Substitue i^2=-1
\displaystyle =\displaystyle 4x^2-12ix-9Evalute the multiplication

This was the original expression, so our factorization is correct.

Example 4

Verify the identity \left(a+bi\right)^4=\left(\left(a^2-b^2\right)+2abi\right)^2.

Worked Solution
Create a strategy

Verifying or showing an identity is less formal than proving the identity. In this problem, we can expand both sides to show both expressions are equal.

Apply the idea

Expanding the left expression first, we get:

\displaystyle \left(a+bi\right)^4\displaystyle =\displaystyle \left(a+bi\right)^2\left(a+bi\right)^2Expand the power
\displaystyle =\displaystyle \left(a^2+2abi+b^2i^2\right)\left(a^2+2abi+b^2i^2\right)Perfect square trinomial identity
\displaystyle =\displaystyle \left(a^2+2abi-b^2\right)\left(a^2+2abi-b^2\right)Substitute i^2=-1
\displaystyle =\displaystyle a^4+2a^3bi-a^2b^2+2a^3bi+4a^2b^2i^2\\-2ab^3i-a^2b^2-2ab^3i+b^4Distributive property
\displaystyle =\displaystyle a^4+2a^3bi-a^2b^2+2a^3bi-4a^2b^2\\-2ab^3i-a^2b^2-2ab^3i+b^4Substitute i^2=-1
\displaystyle =\displaystyle a^4-6a^2b^2+b^4+4a^3bi-4ab^3iCombine like terms

In this last step, we collected the real terms and collected the imaginary terms together, then factored out the i to express it as a complex number in the form a+bi where a represents the real parts and bi represents the imaginary parts.

Next, we expand the expression on the right side:

\displaystyle \left(a^2-b^2+2abi\right)^2\displaystyle =\displaystyle \left(a^2-b^2+2abi\right)\left(a^2-b^2+2abi\right)Expand the power
\displaystyle =\displaystyle a^4-a^2b^2+2a^3bi-a^2b^2+b^4\\-2ab^3i+2a^3bi-2ab^3i+4a^2b^2i^2Distributive property
\displaystyle =\displaystyle a^4-a^2b^2+2a^3bi-a^2b^2+b^4\\-2ab^3i+2a^3bi-2ab^3i-4a^2b^2Substitute i^2=-1
\displaystyle =\displaystyle a^4-6a^2b^2+b^4+4a^3bi-4ab^3iCombine like terms

We found both sides expanded to the same expression, so we have verified the identity.

Reflect and check

In this last step, we could have collected the real terms and collected the imaginary terms together, then factored out the i.

\left(a^4-6a^2b^2+b^4\right)+\left(4a^3b-4ab^3\right)i

Doing this expresses the answer as a complex number in the form a+bi where a represents the real parts and bi represents the imaginary parts.

Idea summary

Important identities we use often are:

  • Perfect square trinomials: a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}
  • Difference of two squares: a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)
  • Sum of two cubes: a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)
  • Difference of two cubes: a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)

We can verify identities mathematically through algebraic manipulation or using geometric diagrams. These identities can be help us describe numerical relationships, and they can be extended to complex numbers.

Outcomes

N.CN.C.8 (+)

Extend polynomial identities to the complex numbers.

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.C.4

Prove polynomial identities and use them to describe numerical relationships.

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