Examples of Polynomial Equations

Solving polynomial equations of degree larger than two can be difficult, and in most cases the only way to do it is to have a computer search for them. There are formulas for degree 3 and 4, but they are complicated, and there are no formulas for above degree 4. But for certain polynomials, particularly those with some rational solutions, there are ways to solve them by hand. The following three theorems can be helpful for this.

• The Rational Zero Theorem
• The Upper and Lower Bound Theorem
• Descartes Rule of Signs

The Rational Zero Theorem

If P(x) has integer coefficients, and p/q is a rational zero of P(x), then p is a factor of the leading coefficient (the coefficient on the highest power) and q is a factor of the constant coefficient.

The Upper and Lower Bound Theorem

If a>0 and all of the numbers in the bottom  row of the synthetic division of P divided by x-a are positive, then a is an upper bound for the zeros of P.

If a<0 and the numbers in the bottom row of the synthetic division of P divided by x-a alternate signs, then a is a lower bound for the zeros of P.

Descartes Rule of Signs

Write the polynomial P in standard form, that is with the terms in order of descending powers.

The number of positive zeros of P counting multiplicity is equal to the number of changes of signs of P or that number decreased by an even integer.

The number of negative zeros of P counting multiplicity is equal to the number of positive zeros of P(-x), so it is the number of changes of signs of P(-x) or that number decreased by an even integer.

Examples

Here are some  examples of how to use these theorems. I made up these examples by working backwards from the solutions that I wanted, but other than that I have tried to be honest and show the working of the examples is close as possible to the way I would think if I were approaching the problems. We all think differently, so this in no way means that you have to do them the same way, but I hope it will give you a useful example of one person's way of thinking. For a more formal approach see pretty much any College Algebra text. College Algebra and Trigonometry by Aufmann, Barker, and Nation is the text that we use is the text that we use, but there are plenty of other good ones.

Example 1

Solve the equation.

x⁴+14x³+71x²+154x+120=0

The Rational Zero Theorem tells me that if a solution is p/q that q must be a factor of 1, and p must be a factor of 120. So here are the possibilities.

Numerator: ±1,2,3,4,5,6,8,10,12,15,20,24,30,60,120
Denominator : ±1
Possible rational solutions : ±1,2,3,4,5,6,8,10,12,15,20,24,30,60,120

(Actually in a problem like this, I might not list out all the possibilities at the beginning, because I would figure that particularly if it was a problem from a textbook or on a test that I would probably find enough solution before getting up to the larger numbers. I would check the smaller numbers and then list more numbers only if I had to, but I have listed them all at the beginning, because it may be neater and easier to follow that way for those who are new at it.)

There are no changes of signs, so by Descartes Rule of Signs, there are no positive solutions. P(-x) has 4 changes of of signs, so Descartes Rule of Signs doesn't much restriction on the number of negative solutions. That leaves a lot of possibilities, but hopefully we will come up with enough solutions before going through all of them. Start with the easiest, which is -1.

so we can see that it doesn't work. It is nicest to do these on a white board, so if you can get a hold of one you should try it. If I was at the board right now I could just erase the -1 and the bottom two rows, but leave up the coefficients and try another number. -2 is the next easiest number to try. It works!

Having found one solution we can reduce the degree of our polynomial by one, because from the division we get that our polynomial factors to (x+2)(x³+12x²+47x+60). By the principle of zero factors the only way this can be zero is if one of the factors is zero. So to find other solutions besides -2, we merely need to find solutions to the second factor. A convenient way I have found to do that is to cross out the 0 remainder at the end and just use the bottom row as the coefficients of my new polynomials. The first thing to check is to see if -2 is still a zero for this polynomial.

No, it doesn't work, but we can see that there could still be a smaller (larger negative) solution, because the signs aren't alternating on the bottom. (Upper and Lower Bound Theorem) But of course there is no guarantee that this will be one of ours, because it could be irrational. But no harm in trying -3, so again we can erase and try it. And we are in luck again, because it works. This gives us that our polynomial factors into (x+2)(x+3)(x²+9x+20), so again by the principle of zero factors we get that either x+2=0, x+3=0, or x²+9x+20=0. The first two we have already taken care of. For the third factor, we could continue with the synthetic division, but we don't have to, because this is a quadratic, and quadratics are much easier to solve. If we can't figure out how to factor it, we can use the quadratic formula. Let's see if it factors. Can we find two numbers whose product is 20 and whose sum is 9? Yes, we can, 4 and 5. So it factors into (x+4)(x+5) and the whole polynomial factors into (x+2)(x+3)(x+4)(x+5) and the solutions to the equation are x=-2,-3,-4,-5.

Example 2

Solve the equation.

2x⁴+3x³-16x²+15x-4=0

Numerator: ±1,2,4
Denominator: ±1,2
Possible rational solutions: ±1,2,4,1/2

If 1 is a possibility I always try it first because I am lazy, and it is easiest. Since there are changes of signs this time there is no reason to exclude it.

And this time we are in luck and it works. We might as well be really lazy and try it again, and as before we can just continue on with the new coefficients at the bottom, because of the principle of zero products. And what do you know, it worked again. So now we are home free, because we have reduced it down to a quadratic again. The original polynomial then factors to (x-1)²(2x²+7x-4), so to finish the solution we just need to solve 2x²+7x-4=0. Let's see if this one factors. To factor it we need something of the form

(2x          )(x          ).

If we use 4 and 1 and put the 4 so that it multiplies by the 2 and the 1 so that it multiplies by 1, then we get 8-1=7 for the middle coefficient, so that works, and we get (2x-1)(x+4), so the original polynomial factors to (x-1)²(2x-1)(x+4), giving us solutions of 1,1/2, and -4. But wait a minute, there's something funny here. There are 3 changes of signs in the original polynomials, so by Descartes Rule of Signs there should be 3 or 1 positive solution and there are 2. What's wrong? To figure out what's wrong with have to look back at Descartes Rule of Signs and see that it says "counting multiplicity". Since x-1 is a factor twice, 1 is a zero of multiplicity 2, so counting muliplicity, there are 3 positive zeros.

Example 3

Solve the equation. 15x⁵+37x⁴+97x³+119x²-88x+12=0

Numerator: ±1,2,3,4,6,12
Denominator: ±1,3,5,15
Possible rational solutions: ±1,2,3,4,6,12,1/3,2/3,4/3,1/5,2/5,3/5,4/5,6/5,12/5,1/15,2/15,4/15

Try 1.

It is already clear from this much that not only is 1 not a solution but that 1 is an upper bound on the solutions, so it is not necessary to check 2,3,4,6, or 12. Since fractions are a pain, it is probably best to start with the negative numbers. According to Descartes rule of Signs there are 2 or 0 positive solutions, but even if there are 2 there is no guarantee that they will be rational. Try -1. It doesn't work. Try -2. It works. Will it work again? No, but from here we can tell that -2 is a lower bound, since the signs will alternate. We can tell without doing the addition that -47+-258 is going to be negative and when you multiply that negative by -2 it will be positive and adding that to 6 that will be positive. So there is no point in checking any more negative whole numbers, so we are faced with having to check fractions. Probably the easiest to try is 1/3, so let's erase the second -2 and try it. It works. Try it again. It doesn't work, but it is clear from here that it is an upper bound since even 90 times 1/3 is bigger than 18, so the bottom row numbers will all be positive. So there is no point in trying 2/3 or anything else bigger than 1/3. Probably the best thing to check now is 1/5. Eureka! It works, and now we have it reduced to a quadratic, so we can solve it. The original polynomial then factors to (x+2)(x-1/3)(x-1/5)(15x²+15x+90). Factoring out 15, this last factor becomes

x²+x+6.

This doesn't factor, but we can use the quadratic formula to solve it.

b²-4ac=1-4(6)=-23

so

All together then the solutions to the equations are

.

Example 4

Solve the equation. x⁶+ 5x⁵+ x⁴ -25x³ -24x²+ 30x+ 36=0

Numerator: ±1,2,3,4,6,9,12,18
Denominator: ±1
Possible rational solutions:  ±1,2,3,4,6,9,12,18

By adding up the coefficients we can see that 1 won't work. There are two changes of signs, so Descartes Rule of Signs says that there could be 2 positive solutions, so let's try 2.

2 doesn't work and it also doesn't show 2 as an upper bound, so how about 3. It is clear from here that it is not going to work and furthermore that it is an upper bound to the solutions, since 3 times 50 will be much bigger than 24 and from then all everything will be positive. So it looks like there aren't going to be any positive rational solutions. According to Descartes Rule of Signs there are two possibilites here. There could be indeed be no positive solutions, because the number of solutions can differ be a multiple of two from the number of changes of signs, but it also could be that there are 2 irrational positive solutions. In either case we don't have any tools to help us find them, so it is time to concentrate on the negative solutions. Try -1. Nope, it doesn't work. Close. Maybe -2 will work. Yes, we've got it. 1 down, 5 to go. :-( Oh well, we must perservere. Will it work again? No such luck. How about -3? Yes!!! And look at the interesting polynomial we get when dividing it out. Those zeros are going to make it much easier to find the remaining solutions. Our original polynomial factors into (x+2)(x+3)(x⁴-5x²+6). This 3rd polynomial is sort of a quadratic in disguise. It can be factored by treating x² like it were the variable and factoring the same way you would factor x²-5x+6, find two numbers whose product is 6 and whose sum is -5. The two numbers are -2 and -3, so it factors to (x²-2)(x²-3). Setting these equal to 0 we get as solutions, so all together the solutions to the original equation are