The rational root theorem says that, if you have a polynomial equation written in the form , then you can make a list of all the possible rational roots by looking at the first term and the last term. Any rational roots must be able to be written as a fraction with a factor of the constant (the last term or a0) in the numerator of the fraction and a factor of the lead coefficient (an) in the denominator.
For example, in the equation 4x4 – 3x3 + 5x2 + 9x – 3 = 0, the factors of the constant are +3, –3, +1, –1 and the factors of the coefficient of the first term are +4, –4, +2, –2, +1, –1. The following list includes all the ways that you can create a fraction with a factor of the constant in the numerator and a factor of the lead coefficient in the denominator:
Of course, the two fractions with 1 in the denominator are actually whole numbers, when you simplify. This abbreviated listing represents all the possible ways to combine +3 and –3 and +4 and –4, and so on, to create all the possible fractions. It’s just quicker to use the ± notation than to write out every single possibility.
Although this new list has 12 candidates for solutions to the equation, it’s really relatively short when you’re trying to run through all the possibilities. Many of the polynomials start out with a 1 as the coefficient of the first term, which is great news when you’re writing your list because that means the only rational numbers you’re considering are whole numbers — the denominators are 1.
Q. Determine all the possible rational solutions of this equation: 2x6 – 4x3 + 5x2 + x – 30 = 0.
A.
The factors of the constant are and the factors of the lead coefficient are . You create the list of all the numbers that could be considered for roots of the equation by dividing each of the factors of the constant by the factors of the lead coefficient. The numbers shown in the answer don’t include repeats or unfactored fractions.
Q. Determine all the possible rational solutions of the equation: x6 – x3 + x2 + x – 1 = 0.
A.
Yes, even though Descartes tells you that there could be as many as three positive and one negative real root, the only possible rational roots are +1 or –1.
3. List all the possible rational roots of 2x4 – 3x3 – 54x + 81 = 0.
Solve It
4. List all the possible rational roots of 8x5 – 25x3 – x2 + 25 = 0.
Solve It
Using the Factor/Root Theorem
Algebra has a theorem that says if the binomial x – c is a factor of a polynomial (it divides the polynomial evenly, with no remainder), then c is a root or solution of the polynomial. You may say, “Okay, so what?” Well, this property means that you can use the very efficient method of synthetic division to solve for solutions of polynomial equations.
Use synthetic division to try out all those rational numbers that you listed as possibilities for roots of a polynomial. (See Chapter 8 for more on synthetic division.) If x – c is a factor (and c is a root), then you won’t have a remainder (the remainder is 0) when you perform synthetic division.
Q. Check to see whether the number 2 is a root of the following polynomial:
x6 – 6x5 + 8x4 + 2x3 – x2 – 7x + 2 = 0
A. Yes, it’s a root. Use the 2 and the coefficients of the polynomial in a synthetic division problem:
The remainder is 0, so x – 2 is a factor, and 2 is a root or solution. The quotient of this division is x5 – 4x4 + 2x2 + 3x – 1, which you write using the coefficients along the bottom. When writing the factorization, make sure you start with a variable that’s one degree lower than the one that was divided into. This new polynomial ends in a –1 and has a lead coefficient of 1, so the only possible solutions when setting the quotient equal to zero are 1 or –1.
Q. Check to see whether 1 or –1 is a solution of the new equation.
A. Neither is a solution. First, try 1:
That 1 didn’t work; the remainder isn’t 0. Now, try –1:
It doesn’t work, either. The only rational solution of the original equation is 2.
5. Check to see whether –3 is a root of x4 – 10x2 + 9 = 0.
Solve It
6. Check to see whether is a root of 2x4 – 3x3 – 54x + 81 = 0.
Solve It
Solving By Factoring
When determining the solutions for polynomials, many techniques are available to help you determine what those solutions are — if any solutions exist. One method that is usually the quickest, though, is factoring and using the multiplication property of zero (MPZ). (Check out Chapter 13 for more ways to use the MPZ.) Not all polynomials lend themselves to factoring, but when they do, using this method is to your advantage. And don’t forget to try synthetic division and the factor theorem if you think you have a potential solution.
Q. Find the real solutions of x4 – 81 = 0.
A. x = 3 or x = –3. Do you recognize that the two numbers are both perfect squares? Factoring the binomial into the sum and difference of the roots, you get (x2 – 9)(x2 + 9) = 0. The first factor of this factored form is also the difference of perfect squares. Factoring again, you get (x – 3)(x + 3)(x2 + 9) = 0. Now, to use the MPZ, set the first factor equal to 0 to get x – 3 = 0, x = 3. Set the second factor equal to 0, x + 3 = 0, x = –3. The last factor doesn’t cooperate: x2 + 9 = 0, x2 = –9. A perfect square can’t be negative, so this factor has no solution. If you go back to the original equation and use Descartes’s rule of signs (see “Determining How Many Possible Roots” earlier in this chapter), you see that it has one real positive root and one real negative root, which just confirms that prediction.
Q. Find the real solutions of x4 + 2x3 – 125x – 250 = 0.
A. x = –2 or x = 5. You can factor by grouping to get x3(x + 2) – 125(x + 2) = (x + 2)(x3 – 125) = 0. The second factor is the difference of perfect cubes, which factors into (x + 2)(x – 5)(x2 + 5x + 25) = 0. The trinomial factor in this factorization never factors any more, so only the first two factors yield solutions: x + 2 = 0, x = –2 and x – 5 = 0, x = 5.
But suppose that, instead of factoring, you make a guess (after determining with Descartes’s rule of signs that only one positive real root and three or one negative real roots are possible). You go with a negative guess (go with the odds). How about trying –2?
The –2 is a solution (no surprise there), and the factored form is . The difference of cubes gives you that same solution of x = 5. You can use both synthetic division and factoring in the same problem!
7. Solve by factoring: x4 – 16 = 0.
Solve It
8. Solve by factoring: .
Solve It
9. Solve by factoring: x3 + 5x2 – 16x – 80 = 0.
Solve It
10. Solve by factoring: x6 – 9x4 – 16x2 + 144 = 0.
Solve It
Solving Powers That Are Quadratic-Like
A special classification of equations, called quadratic-like, lends itself to solving by factoring using unFOIL. These equations have three terms:
A variable term with a particular power
Another variable with a power half that of the first term
A constant
A way to generalize these characteristics is with the equation ax2n + bxn + c = 0. In Chapter 11, you’ll find details on factoring these quadratic-like trinomials, which then allows you to solve the related equations.
Q. Solve for x: x8 – 17x4 + 16 = 0.
A. x = 1 or x = –1 or x = 2 or x = –2
1. Factor the expression:
(x4 – 1)(x4 – 16) = 0.
2. Factor each of the binomials.
You can factor x4 – 1 into (x2 – 1)(x2 + 1). You can factor x4 – 16 into (x2 – 4)(x2 + 4).
3. Factor the binomials that are still the difference of perfect squares.
(x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x2 + 1)
x4 – 16 = (x – 2)(x + 2)(x2 + 4)
4. Set the complete factorization equal to 0.
(x – 1)(x + 1)(x2 + 1)(x – 2)(x + 2)(x2 + 4) = 0
The four real
solutions of the original equation are x = 1, –1, 2 and –2. The two factors that are the sums of perfect squares don’t provide any real roots.
Q. Solve for y: y2/3 + 5y1/3 + 6 = 0.
A. y = –8 or y = –27. This example involves fractional exponents. Notice that the power on the first variable is twice that of the second.
1. Factor the expression.
(y1/3 + 2)(y1/3 + 3) = 0.
2. Take the first factor and set it equal to 0.
y1/3 + 2 = 0
3. Add –2 to each side and cube each side of the equation.
4. Follow the same steps with the other factor.
11. Solve the equation: x4 – 13x2 + 36 = 0.
Solve It
12. Solve the equation: x10 – 31x5 – 32 = 0.
Solve It
13. Solve the equation: y4/3 – 17y2/3 + 16 = 0.
Solve It
14. Solve the equation: z–2 + z–1 – 12 = 0.
Solve It
Answers to Problems on Solving Higher Power Equations
This section provides the answers (in bold) to the practice problems in this chapter.
1. Count the number of possible positive and negative real roots in . The answer: Three or one positive roots and two or no negative roots. The original equation has three sign changes, so there are three or one possible positive real roots. Then substituting –x for each x, you get , and you have two sign changes, meaning there are two negative roots or none at all.
2. Count the number of possible positive and negative real roots in . The answer: Two or no positive roots and one negative root. The original equation has two sign changes (from positive to negative to positive), so there are two or no positive roots. Substituting –x for each x, you get , which has one sign change and one negative root.
3. List all the possible rational roots of . The answer: The possible rational roots are . The constant term is 81. Its factors are . The lead coefficient is 2 with factors .
4. List all the possible rational roots of . The possible rational roots are . The constant term is 25, having factors . The lead coefficient is 8 with factors .
5. Check to see whether –3 is a root of . The answer: Yes. Rewrite the equation with the coefficients showing in front of the variables: .
Because the remainder is 0, the equation has a root of –3 and a factor of (x + 3).
6. Check to see whether is a root of . The answer: Yes. Writing in the coefficients of the terms, you get . Note that 3 is a factor of 81 and 2 is a factor of 2, so is a possible rational root:
The remainder is 0, so is a root (solution), and or (2x – 3) is a factor.
7. Solve by factoring: . The answer is . First factor the binomial as the difference and sum of the same two values; then factor the first of these factors the same way:
are the real solutions. So . Note: has no real solutions.
8. Solve by factoring: . The answer is . First factor out x from each term to get . You may recognize that the expression in the parentheses is a perfect cube. But just in case that hasn’t occurred to you, try using synthetic division with the guess x = 1.
The factored form now reads . The trinomial in the parentheses is a perfect square, so the final factored form is . So , and 1 is a triple root.
9. Solve by factoring: . The answer is . Factor by grouping:
So x – 4 = 0, x = 4; x + 4 = 0, x = –4; or x + 5 = 0, x = –5. Therefore, .
10. Solve by factoring: . The answer is . First you get by grouping. Then . So give the real solutions. Therefore, .
11. Solve the equation: . The answer is .
You can continue factoring or use the square root rule.
. Therefore .
12. Solve the equation: . The answer is .
So . Therefore, .
13. Solve the equation: . The answer is .
, so
or . Therefore, y = 1 or y = 64.
14. Solve the equation: . The answer is .
or
Therefore, .
Chapter 15
Reeling in Radical and Absolute Value Equations
In This Chapter
Squaring radicals once or twice (and making them nice)
Being absolutely sure of absolute value
Checking for extraneous solutions
Radical equations and absolute values are just what their names suggest. Radical equations contain one or more radicals (square root or other root symbols), and absolute value equations have an absolute value operation (two vertical bars that say to give the distance from 0). Although they’re two completely different types of equations, radical equations and absolute value equations do have something in common: You change both into linear or quadratic equations (see Chapters 12 and 13) and then solve them. After all, going back to something familiar makes more sense than trying to develop (and then remember) a bunch of new rules and procedures.
What’s different is how you change these two types of equations. I handle each type separately in this chapter and offer practice problems for you.
Squaring Both Sides to Solve Radical Equations
If your radical equation has just one radical term, then you solve it by isolating that radical term on one side of the equation and the other terms on the opposite side, and then squaring both sides. After solving the resulting equation, be sure to check your answer or answers to be sure you actually have a solution.
Watch out for extraneous roots. These false answers crop up in several mathematical situations where you change from one type of equation to another in the course of solving the original equation. In this case, it’s the squaring that can introduce extraneous or false roots. Creating these false roots happens because the square of a positive number or its opposite (negative) gives you the same positive number.
For example, if you start with the simple statement that x = 2 and then square both sides, you get x2 = 4. Now take the square root of both sides, and you get x = ±2. All of a sudden you have an extraneous root because the –2 got included (it wasn’t part of the original statement).
Squaring both sides to get these false answers may sound like more trouble than it’s worth, but this procedure is still much easier than anything else. You really can’t avoid the extraneous roots; just be aware that they can occur so you don’t include them in your answer.
When squaring both sides in radical equations, you can encounter one of three possible outcomes: (1) both of the solutions work, (2) neither solution works. or (3) just one of the two solutions works because the other solution turns out to be extraneous.
Check out the following examples to see how to handle an extraneous root in a radical equation.
Q. Solve for x:
A. x = 6
1. Isolate the radical term by subtracting x from each side. Then square both sides of the equation.
2. To solve this quadratic equation, subtract x and 10 from each side so that the equation is set equal to zero. Then simplify and then factor it.
Two solutions appear, x = 15 or x = 6.
3. Check for an extraneous solution.
In this case, substituting the solutions, you see that the 6 works, but the 15 doesn’t.
Q. Solve for x:
A. x = –2 or x = –3. It’s unusual to have both answers from the quadratic work in the original radical equation.
Algebra I Workbook For Dummies Page 13