Zeros of a Polynomial Function Lesson

Over the course of the last few lessons, we have been discussing various tools and techniques that can be used to find the zeros for a polynomial function. Before moving forward with this tutorial, you should have an understanding of the Factor Theorem, the Rational Zeros Theorem, Descartes' Rule of Signs, and the Upper and Lower Bounds Theorem. Finding the zeros of a polynomial function is one of the more tedious problem types in a College Algebra or Precalculus class. Keep in mind that some of these problems are going to take a good bit of time.

Steps for Finding Zeros

1. Try to factor the polynomial function
   • In easier cases, the zeros can be found by simply factoring
2. Use Descartes' Rule of Signs
   • This will tell us the possible number of positive and negative real zeros
3. Use the Rational Zeros Theorem to list all possible rational zeros
4. Test the potential rational zeros using synthetic division (Remainder Theorem)
5. Use the Upper and Lower Bounds Theorem to narrow down the possibilities
   • When searching for positive real zeros, start with the smallest positive number on the list
     • If the bottom row of the synthetic division contains only non-negative entries, then you will not have any real zeros greater than the number that was tested
   • When searching for negative real zeros, start with the greatest negative number on the list (the negative number that is closest to zero)
     • If the bottom row of the synthetic division alternates in sign (where 0 can be used as positive or negative as needed), then you will not have any real zeros less than the number that was tested
   • The Upper and Lower Bounds Theorem is a one-sided test
     • If a number passes the test, then it is definitely an upper/lower bound
     • If a number fails the test, we can't determine if it is an upper/lower bound
6. Once a zero is found, use the Factor Theorem to rewrite the polynomial function
   • f(x) = (x - k)q(x), where k is the zero we found
   • If q(x) is a quadratic we can stop and use the quadratic formula
   • If q(x) is not a quadratic:
     • Try to factor q(x)
     • Test the remaining candidates on q(x)
     • The zero you just found should be retested on q(x), you may have a zero of multiplicity
     • Candidates that were not zeros for f(x) will not be zeros for q(x) and don't need to be retested

Keep in mind that the above list is only going to work for the typical problems given in this section. If you have a problem that has a degree of 3 or higher, where we can't factor or there are no rational zeros, this strategy is not going to work and more advanced techniques are needed. Additionally, this is just a suggested order. You might find that for a particular problem, a different order or skipping certain items could be faster. Let's look at an example.
Example #1: Find all zeros. $$f(x)=x^3 - 3x^2 - 12x + 10$$ First, check to see if you can factor the polynomial as it stands. In this case, we have a four-term polynomial. We would try factoring by grouping but that doesn't work here. Let's move on and think about a few things.

• From the Fundamental Theorem of Algebra, we know that we have at most 3 distinct solutions
  • We will have exactly 3 zeros when a zero of multiplicity is taken into account
• From Descartes' Rule of Signs, we obtain a list of possible positive and negative real roots:
  • # of positive real zeros: 2 or 0
    • $$f(x)=x^3 - 3x^2 - 12x + 10$$
    • There are 2 sign changes in f(x)
  • # of negative real zeros: 1
    • $$f(-x) = -x^3 - 3x^2 + 12x + 10$$
    • There is only 1 sign change in f(-x)
• From the rational roots test, we obtain a list of possible rational roots:
  • $$\pm 1, \pm 2, \pm 5, \pm 10$$

From here, we can use synthetic division, along with our upper and lower bound rules to narrow down the possibilities. Since we have a third-degree polynomial, we know that we only have to find one rational zero and then we will be able to use the quadratic formula to find the remaining two zeros. From Descartes' Rule of Signs, we know that there is exactly 1 negative real zero. The issue is that we don't know if it's rational or irrational. This means the negative real zero would be on the list if it's rational or not be on the list if it's irrational.
Let's begin by testing -1 since this is the largest negative number on the list (closest to zero). If the number is not a zero, then we can use the lower bound test to see if we need to keep going lower and testing other negative numbers on the list. Here, we can conclude that -1 is not a zero. The lower bound test would be inconclusive here as the signs don't alternate (remember this is a one-sided test). Now we will test -2 since it is the next largest negative number on the list. Again, we find that -2 is not a zero and the lower bound test is inconclusive (one-sided test). Now we will test -5. Although we find that -5 is not a zero, the lower bound test shows conclusively that we will not have a real zero that is less than -5. This means we do not have to check -10 and our negative real zero is going to be irrational. Now we will check the positive rational zeros from the list. From Descartes' Rule of Signs, we know there are either 2 or 0 positive real zeros. Let's start with 1 since it is the smallest positive number on the list. Here, we can see that 1 is not a zero and the upper bound test is inconclusive (one-sided test) as the bottom row would need to be all non-negative entries to show an upper bound. Let's move up to 2 and see if that works. Again, 2 is not a zero and the upper bound test is inconclusive (one-sided test). Let's now try 5 and see if that works. We can see that 5 is a zero and the upper bound test is inconclusive (one-sided test). Let's set up the quotient using the results of the synthetic division. $$\frac{x^3-3x^2-12x+10}{x-5}=x^2 + 2x - 2$$ We can use this to factor our polynomial: $$f(x)=(x - 5)(x^2 + 2x - 2)$$ Let's use the quadratic formula to solve for the remaining zeros. $$x^2 + 2x - 2 = 0$$ $$a = 1, b = 2, c = -2$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-2 \pm \sqrt{(2)^2 + (-4)(1)(-2)}}{2(1)}$$ $$x = \frac{-2 \pm \sqrt{4 + 8}}{2}$$ $$x = \frac{-2 \pm \sqrt{12}}{2}$$ $$x = \frac{-2 \pm 2\sqrt{3}}{2}$$ $$\require{cancel}x = \frac{\cancel{2}(-1 \pm \sqrt{3})}{\cancel{2}}$$ $$x = -1 \pm \sqrt{3}$$ Putting everything together, we can say that our polynomial function has the following zeros: $$-1 + \sqrt{3}, -1 - \sqrt{3}, 5$$ Alternatively, we could write our zeros in this way: $$(x - 5)(x^2 + 2x - 2) = 0$$ $$x = -1 + \sqrt{3}, -1 - \sqrt{3}, 5$$ As you can see, that was quite a bit of work to solve one problem. Some problems are easier (less tedious) and others are harder (more tedious). Generally the larger the degree, the more time things take as you have more steps. Also, having a larger list of potential zeros quickly adds additional time to the process. Let's look at one more example.
Example #2: Find all zeros. $$f(x)=x^5 - x^4 + 2x^3 - 2x^2 - 3x + 3$$ First, check to see if you can factor the polynomial as it stands. In this case, we can actually factor. $$f(x)=x^5 - x^4 + 2x^3 - 2x^2 - 3x + 3$$ $$=x^4(x - 1) + 2x^2(x - 1) -3(x - 1)$$ $$=(x - 1)(x^4 + 2x^2 - 3)$$ $$=(x - 1)(x^2 - 1)(x^2 + 3)$$ $$=(x - 1)^2(x + 1)(x^2 + 3)$$ At this point, we know we would have a zero of 1 with multiplicity 2, and then a zero of -1. We would only need to figure out the quadratic. $$x^2 + 3 = 0$$ $$x^2 = -3$$ $$x = \pm \sqrt{-3} = \pm i\sqrt{3}$$ So the zeros would be: $$1 (\text{mult. 2}), -1, i\sqrt{3}, -i\sqrt{3}$$ The factoring shown above is probably not obvious, so let's also show how we could have gotten this using our steps.

• From the Fundamental Theorem of Algebra, we know that we have at most 5 distinct solutions
  • We will have exactly 5 zeros when a zero of multiplicity is taken into account
• From Descartes' Rule of Signs, we obtain a list of possible positive and negative real roots:
  • # of positive real zeros: 4, 2, or 0
    • $$f(x)=x^5 - x^4 + 2x^3 - 2x^2 - 3x + 3$$
    • There are 4 sign changes in f(x)
  • # of negative real zeros: 1
    • $$f(-x) = -x^5 - x^4 - 2x^3 - 2x^2 + 3x + 3$$
    • There is only 1 sign change in f(-x)
• From the rational roots test, we obtain a list of possible rational roots:
  • $$\pm 1, \pm 3$$

From here, we can use synthetic division, along with our upper and lower bound rules to narrow down the possibilities. From Descartes' Rule of Signs, we know that there is exactly 1 negative real zero. The issue is that we don't know if it's rational or irrational. This means the negative real zero would be on the list if it's rational or not be on the list if it's irrational.
Let's begin by testing -1 since this is the largest negative number on the list (closest to zero). If the number is not a zero, then we can use the lower bound test to see if we need to keep going lower and testing other negative numbers on the list. We can see that -1 is a zero and a lower bound. Let's set up the quotient using the results of the synthetic division. $$\frac{x^5 - x^4 + 2x^3 - 2x^2 - 3x + 3}{x + 1} = x^4 - 2x^3 + 4x^2 - 6x + 3$$ We can use this to factor our polynomial: $$(x + 1)(x^4 - 2x^3 + 4x^2 - 6x + 3)$$ Now, we will work on the quotient: $$x^4 - 2x^3 + 4x^2 - 6x + 3$$ At this point, you could factor. This is something you should always try as you may have missed the factoring at the beginning. $$(x^4 + 4x^2 + 3) + (-2x^3 - 6x)$$ $$=(x^2 + 3)(x^2 + 1) -2x(x^2 + 3)$$ $$=(x^2 + 3)(x^2 - 2x + 1)$$ $$=(x^2 + 3)(x - 1)^2$$ Once again, it may not be obvious that you can factor, so we will proceed using the steps for the sake of the tutorial.
From Descartes' Rule of Signs, we know that we only have 1 negative real zero. This means as we return to our list, we will only work with positive numbers. Let's now try 1. We can see that 1 is a zero but the upper bound test is inconclusive (one-sided test). Let's set up the quotient using the results of the synthetic division. $$\frac{x^4 - 2x^3 + 4x^2 - 6x + 3}{x - 1} = x^3 - x^2 + 3x - 3$$ We can use this to factor our polynomial: $$(x + 1)(x - 1)(x^3 - x^2 + 3x - 3)$$ Now, we will work on the quotient: $$x^3 - x^2 + 3x - 3$$ Note, this can be factored using grouping: $$x^2(x - 1) + 3(x - 1)$$ $$(x - 1)(x^2 + 3)$$ Again, for the sake of the tutorial, we will also show how we could have gotten to this point using synthetic division. We would need to check 1 in the quotient since we may have a zero of multiplicity. We can see that 1 is a zero and an upper bound. Let's set up the quotient using the results of the synthetic division. $$\frac{x^3 - x^2 + 3x - 3}{x - 1} = x^2 + 3$$ We can use this to factor our polynomial: $$(x + 1)(x - 1)^2(x^2 + 3)$$ Now we can get the final zeros using the quadratic formula but in this situation, it's not necessary as we can just use the square root property. $$x^2 + 3 = 0$$ $$x^2 = -3$$ $$x = \pm \sqrt{-3}$$ $$x = \pm i\sqrt{3}$$ Putting everything together, we can say that our polynomial function has the following zeros: $$-1, 1 (\text{mult.} \, 2), i\sqrt{3}, -i\sqrt{3}$$ Alternatively, we could write our zeros in this way: $$(x + 1)(x - 1)^2(x^2 + 3) = 0$$ $$x = -1, 1, i\sqrt{3}, -i\sqrt{3}$$

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