Tutorial: Rational Root Theorem and Factoring Polynomials

Now that we’ve got an idea about how to sketch factored polynomials, taking into account the end behavior and the crosses, bounces, and slides at the x-axis, let’s figure out how to deal with polynomials that start out in standard (un-factored) form. We’ll focus on the Rational Root Theorem as our method of choice, which, as you may have guessed from the name, will help find any rational roots the polynomial might have (rational: numbers that can be written as a fraction of whole numbers).

All of the Possibilities

The Rational Root Theorem lets us find all of the rational numbers that could possibly be roots of the equation. Sometimes the list of possibilities we generate will be big, but it’s still a finite list, so it’s a better start than randomly trying out numbers to see if they are roots.

Rational Root Theorem: Step By Step

  1. Write down all of the factors of the constant term of the polynomial, including itself and one.
  2. Write down all of the factors of the leading coefficient.
  3. Write down all possible fractions where the numerator is a factor of the constant term, and the denominator is a factor of the leading coefficient. Try to be systematic so that you get all of the possibilities.
  4. These fractions, both the positive and negative version of each, are your possible rational roots.
  5. Now that you have a list of possible roots, you can test each one to see if it is a root of the polynomial. Test either by plugging the number in for x in the polynomial (it’s a root if the result is 0), or by synthetically dividing by the number (it’s a root if the remainder is 0).
  6. Cross possibilities that don’t work off of your list as you test them.
  7. When you find a root that works, synthetically divide by that root to get the lower degree polynomial that results when it’s factored out. Now apply the rational root theorem to this new polynomial – you may have fewer possibilities now!
  8. Once you get down to a quadratic equation, you can solve for the roots using any of the typical quadratic equation methods.

An Example:

Let’s go through the steps with this polynomial: P(x)=2x^3+x^2-13x+6

  1. Constant Term is 6. Factors: 1, 2, 3, 6
  2. Leading Coefficient is 2. Factors: 1, 2
  3. Possible Rational Roots: \frac{1}{1}, \frac{2}{1}, \frac{3}{1}, \frac{6}{1}, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{6}{2}. These simplify to 1, 2, 3, 6, \frac{1}{2}, 1, \frac{3}{2}, 3. Some of those are repeats, so our final list is \pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}; making a total of 12 possibilities.
  4. Test each possible root:
    1. P(1) = 2(1)^3+(1)^2-13(1)+6 = 2+1-13+6=-4 Not a root.
    2. P(-1) = 2(-1)^3+(-1)^2-13(-1)+6 = -2+1+13+6=18 Not a root.
    3. P(2) = 2(2)^3+(2)^2-13(2)+6 = 16+4-26+6=0 We found a root!
  5. Use synthetic or long division to factor out the root: P(x)=(x-2)(2x^2+5x-3)
  6. 2x^2+5x-3 is quadratic so we can find its roots like any other quadratic. I’ll factor: 2x^2+5x-3=(2x-1)(x+3), so P(x)=(x-2)(2x-1)(x+3) The polynomial is factored, and the roots are 2, \frac{1}{2}, and -3.

Another Example:

P(x) = 2x^4-7x^3+3x^2+8x-4

  1. Constant Term is -4. Factors: 1, 2, 4
  2. Leading Coefficient is 1. Factors: 1, 2
  3. Possible Rational Roots: \frac{1}{1}, \frac{2}{1}, \frac{4}{1}, \frac{1}{2}, \frac{2}{2}, \frac{4}{2}. In the end, the possible roots are \pm1, \pm2, \pm4, and \pm\frac{1}{2}, for a total of 8 possible rational roots.
  4. Test Roots (I’ll use synthetic division this time, as shown in the video below!)

So what do you think – can you turn a polynomial from standard form into factored form? Do you have any questions you want to ask me about the process? Please let me know by commenting below!

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