Algebra I Assignment - Exponents, Polynomials, Graphs & Geometry

Instructor: John Hamilton

John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.

The ensuing year-end Algebra I homeschool assignment will challenge students on several math concepts, including how to factor and graph a quadratic equation. This mathematics assignment has been created for the 9th grade student. Updated: 02/27/2021

Assignment Explanation and Topic Overview

When your students think of algebra, including exponents, polynomials, and graphs, the last thing they probably think about is geometry, right? Well, there do exist some subtle connections between the two disciplines.

In addition to what is mentioned above, this synoptic end-of-year assignment will cover topics learned throughout the two freshman Algebra I school semesters, including the distributive property along with the five laws of exponents too.

By the end of this assignment, your students will have completed four steps, solved three problems, and delivered two presentations.

Note - The answers are located at the bottom of the page.

Key Terms

  • Algebra: a branch of mathematics in which letters represent numbers
  • Exponent: a quantity which represents a power to be raised
  • Geometry: a branch of mathematics in which properties of figures are explained


  • Graph paper
  • Online capability
  • Paper
  • Pencil
  • Ruler

Time /Length

  • Two days to complete this Algebra I assignment
  • Two weeks to design an appropriate presentation

Assignment Instructions for Students

Step One

Let's begin! Do you remember the five laws of exponents from earlier this school year? They are:

Product of Powers Rule:

  • xa * xb = xa + b


  • 25 * 26 = 2(5 + 6)

Quotient of Powers Rule:

  • xa / xb = xa - b


  • 38 / 35 = 3(8 - 5)
    • Of course, x doesn't equal zero, because you can't divide by zero.

Power of a Power Property:

  • (xa)b = x(a * b)


  • (42)3 = 4(2 * 3)

Power of a Product:

  • (x * y)a = xa * ya


  • (5 * 6)4 = 54 * 64

Power of a Quotient:

  • (x / y)a = xa / ya


  • (6 / 7)5 = 65 / 75

Step Two

Part I

Problem #1:

Next, let's review what we learned when you want to factor the following quadratic equation:

  • x2 - 6x - 16

Part II

Problem #2:

Now do you recall how to graph this same quadratic equation?

  • f(x) = x2 - 6x - 16

Step Three

Okay, you remember quadratic equations, but do you remember how to solve those ever-so-dastardly cubic equations?

  • x3 - 4x2 + x + 6 = 0

This is a bit tricky, but you can handle it. First, we apply the rational roots theorem or rational roots test. This means all our possible solutions are found by dividing all the factors of the constant (which is 6) by the leading coefficient (the number in front of x3, which in this case is understood to be 1), and include positives and negatives in your list.

  • Factors of 6 = 1, 2, 3, 6
  • Factor of 1 = 1

So our possible solutions are +/- 1, +/-2, +/-3, and +/- 6.

Now I have some bad news, but you may remember we don't have a cool, convenient formula for cubic equations like we do for those quadratic equations. Yep! You actually have to plug in each of these eight possible solutions to find your three solutions. On the other hand, 3 out of 8 aren't horrible odds, right?

Let's start with 1:

  • x3 - 4x2 + x + 6 = 0
  • (1)3 - 4(1)2 + (1) + 6 = 0
  • 1 - 4 + 1 + 6 = 0
  • 4 = 0

Hey, it didn't work. Boo! Hiss! Hiss! Boo! Oh well, let's try again with -1.

  • x3 - 4x2 + x + 6 = 0
  • (-1)3 - 4(-1)2 +(-1) + 6 = 0
  • -1 - 4 - 1 + 6 = 0
  • -6 + 6 = 0
  • 0 = 0

It worked! Since -1 is indeed a zero of the equation, we know (x + 1) is a factor of the equation. Now, to find the other two factors, go ahead and plug in the other 6 numbers. I'll give you 15 minutes.

Are you done?

You should know by this point the three roots, solutions, or zeros are:

  • -1, 2, and 3

Or your three factors are:

  • (x + 1) (x - 2) (x - 3)

By the way, let's recount some of the properties of polynomial functions. First of all, their graphs are continuous. In other words, when drawing one you never lift your pencil from your paper. Second, their graphs are smooth. In other words, your graph will have no sharp corners, but have rounded curves instead.

''How does measurement in algebra relate to geometry though? I thought we already learned algebra and geometry were two completely separate mathematical fields.'' Well, not completely. When you graphed your above polynomial function, you were essentially converting an algebraic concept into a geometric concept.

In a reverse manner, we might be given a graph of a geometric circle, but we use our handy algebra formulas to solve for the perimeter and the area of that circle.

And of course, the Pythagorean theorem combines algebra and geometry concepts.

Step Four

Now let's review the concept of algebraic distribution. You may remember this formula:

  • a(b + c) = ab + ac

We usually read this expression as a times quantity b plus c equals a times b plus a times c.

Problem #3:


  • 5(3x + 2y)


Now it's time to demonstrate your grasp of all the Algebra I material you learned throughout our school year by completing two of the following five presentations:

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