High School Algebra II Unlocked (2016)

Lesson 1.3. Polynomial Identities

REVIEW

PYTHAGOREAN TRIPLE: A set of three positive integers a, b, and c that make the equation a² + b² = c² of the Pythagorean theorem true and therefore represent the side lengths of a right triangle.

VERTEX FORM OF A QUADRATIC EQUATION: y = a(x − h)² + k, with vertex (h, k)

To complete the square for a quadratic equation of the form ax² + bx + c = 0, add a constant to both sides that allows you to rewrite the expression on the left side of the equation as the square of a binomial.

Don’t worry; there are other shortcuts for various polynomial operations that are much simpler than the Binomial Theorem! We call these shortcuts polynomial identities. A polynomial identity is any true equation that shows two equivalent polynomials, often revealing a special pattern that makes the identity more memorable.

To prove a polynomial identity, rewrite the expression on one side of the equation, usually by expanding, and show that it is equivalent to the expression on the other side of the equation.

Prove the polynomial identity (x + y)³ = x³ + y³ + 3xy(x + y).

Putting Polynomial Identities to Use

Polynomial identities are useful beyond simply providing an efficient way to expand basic polynomial products. For example, recognizing patterns from polynomial identities may allow you to solve for certain quantities without necessarily knowing the values of individual variables.

If x + y = 8 and x² + y² = 40, then xy =

A polynomial identity that includes both expressions x + y and x² + y² is the square of a sum, (x + y)² = x² + 2xy + y².

We were able to solve for the value of xy, even though we never specifically solved for the individual values of x and y.

The polynomial identity (x² − y²)² + (2xy)² = (x² + y²)², although not seen as often as those mentioned above, is useful for finding Pythagorean triples.

Find two Pythagorean triples using the polynomial identity above.

This polynomial identity is in the form a² + b² = c², with a = x² − y², b = 2xy, and c = x² + y². We need to choose a value of x and a value of y, then solve for a, b, and c. The absolute value of x must be greater than the absolute value of y, so that x² − y² will be positive. (The value of a, the length of one leg of the triangle, cannot be negative or zero.) Let’s try x = 2 and y = 1.

a = x² − y² = 2² − 1² = 4 − 1 = 3
b = 2xy = 2(2)(1) = 4
c = x² + y² = 2² + 1² = 4 + 1 = 5

One Pythagorean triple is 3-4-5.

That was a very basic Pythagorean triple. Let’s try some slightly larger numbers further apart (although you could certainly use 3 and 1, or 3 and 2, if you wanted). Let x = 5 and y = 2.

a = x² − y² = 5² − 2² = 25 − 4 = 21
b = 2xy = 2(5)(2) = 20
c = x² + y² = 5² + 2² = 25 + 4 = 29

Our second Pythagorean triple is 20-21-29.

Polynomial identities are especially useful for factoring polynomials. Factoring is the key to solving many problems involving polynomial equations.

Ethan threw a ball upward out the window of his apartment. The height of the ball in feet, h, is given by the equation h = −16t² + 48t + 64, where t is the time in seconds since Ethan threw the ball. After how many seconds did the ball hit the ground? After how many seconds did the ball reach its maximum height? What was the maximum height that the ball reached? Draw a graph showing the relationship between time and height.

When the ball hit the ground, h = 0, so we must solve the equation 0 = −16t² + 48t + 64 for t.

The solution t = −1 does not make sense for this situation. The value of t must be positive, as it represents the amount of time since Ethan threw the ball. So, the answer is t = 4. The ball hit the ground 4 seconds after Ethan threw it.

The graph of a quadratic function is a parabola. To find the maximum, the turning point of the parabola, we must put the equation h = −16t² + 48t + 64 in vertex form. To do that, we’ll complete the square.

h = −16(t² − 3t) + 64

Add the square of half the coefficient of t, 3. You must add the same amount to both sides of the equation, so don’t forget to multiply by −16 on the left side.

The vertex is at (3/2, 100), or (1.5, 100). In other words, the ball reached its maximum height of 100 feet 1.5 seconds after Ethan threw it.