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Make sure you aren’t confused by the terminology. All of these are the same: Solving a polynomial equation p(x) = 0 Finding roots of a polynomial equation p(x) = 0 Finding zeroes of a polynomial function p(x) Factoring a polynomial function p(x)There’s a factor for every root, and vice versa. (x−r) is a factor if and only if r is a root. This is theFactor Theorem: inding the roots or inding the factors is essentially the same thing. (The maindifference is how you treat a constant factor.) Exact or Approximate?Most often when we talk about solving an equation or factoring a polynomial, we mean an exact (oranalytic) solution. The other type, approximate (or numeric) solution, is always possible andsometimes is the only possibility. When you can ind it, an exact solution is better. You can always ind a numerical approximationto an exact solution, but going the other way is much more dif icult. This page spends most of its timeon methods for exact solutions, but also tells you what to do when analytic methods fail. Step by StepHow do you ind the factors or zeroes of a polynomial (or the roots of a polynomial equation)?Basically, you whittle. Every time you chip a factor or root off the polynomial, you’re left with apolynomial that is one degree simpler. Use that new reduced polynomial to ind the remaining factorsor roots. At any stage in the procedure, if you get to a cubic or quartic equation (degree 3 or 4), you have achoice of continuing with factoring or using the cubic or quartic formulas. These formulas are a lot ofwork, so most people prefer to keep factoring.Follow this procedure step by step: 1. If solving an equation, put it in standard form with 0 on one side and simplify. [ details ] 2. Know how many roots to expect. [ details ] 3. If you’re down to a linear or quadratic equation (degree 1 or 2), solve by inspection or the quadratic formula. [ details ] Then go to step 7. 4. Find one rational factor or root. This is the hard part, but there are lots of techniques to help you. [ details ] If you can ind a factor or root, continue with step 5 below; if you can’t, go to step 6. 5. Divide by your factor. This leaves you with a new reduced polynomial whose degree is 1 less. [ details ] 2 of 14
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For the rest of the problem, you’ll work with the reduced polynomial and not the original. Continue at step 3. 6. If you can’t ind a factor or root, turn to numerical methods. [ details ] Then go to step 7. 7. If this was an equation to solve, write down the roots. If it was a polynomial to factor, write it in factored form, including any constant factors you took out in step 1.This is an example of an algorithm, a set of steps that will lead to a desired result in a inite number ofoperations. It’s an iterative strategy, because the middle steps are repeated as long as necessary. Cubic and Quartic FormulasThe methods given here— ind a rational root and use synthetic division—are the easiest. But if youcan’t ind a rational root, there are special methods for cubic equations (degree 3) and quarticequations (degree 4), both at Mathworld. An alternative approach is provided by Dick Nickalls for cubicand quartic equations.Step 1. Standard Form and SimplifyThis is an easy step—easy to overlook, unfortunately. If you have a polynomial equation, put all termson one side and 0 on the other. And whether it’s a factoring problem or an equation to solve, put yourpolynomial in standard form, from highest to lowest power. For instance, you cannot solve this equation in this form: x³ + 6x² + 12x = −8You must change it to this form: x³ + 6x² + 12x + 8 = 0Also make sure you have simpli ied, by factoring out any common factors. This may include factoringout a −1 so that the highest power has a positive coef icient. Example: to factor 7 − 6x − 15x² − 2x³begin by putting it in standard form: −2x³ − 15x² − 6x + 7and then factor out the −1 −(2x³ + 15x² + 6x − 7) or (−1)(2x³ + 15x² + 6x − 7)If you’re solving an equation, you can throw away any common constant factor. But if you’re factoring apolynomial, you must keep the common factor. Example: To solve 8x² + 16x + 8 = 0, you can divide left and right by the common factor 8. Theequation x² + 2x + 1 = 0 has the same roots as the original equation. Example: To factor 8x² + 16x + 8 , you recognize the common factor of 8 and rewrite the 3 of 14
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polynomial as 8(x² + 2x + 1), which is identical to the original polynomial. (While it’s true that youwill focus your further factoring efforts on x² + 2x + 1, it would be an error to write that the originalpolynomial equals x² + 2x + 1.)Your “common factor” may be a fraction, because you must factor out any fractions so that thepolynomial has integer coef icients. Example: To solve (1/3)x³ + (3/4)x² − (1/2)x + 5/6 = 0, you recognize the common factor of 1/12and divide both sides by 1/12. This is exactly the same as recognizing and multiplying by the lowestcommon denominator of 12. Either way, you get 4x³ + 9x² − 6x + 10 = 0, which has the same roots asthe original equation. Example: To factor (1/3)x³ + (3/4)x² − (1/2)x + 5/6, you recognize the common factor of 1/12 (orthe lowest common denominator of 12) and factor out 1/12. You get (1/12)(4x³ + 9x² − 6x + 10), whichis identical to the original polynomial.Step 2. How Many Roots?A polynomial of degree n will have n roots, some of which may be multiple roots. How do you know this is true? The Fundamental Theorem of Algebra tells you that thepolynomial has at least one root. The Factor Theorem tells you that if r is a root then (x−r) is a factor.But if you divide a polynomial of degree n by a factor (x−r), whose degree is 1, you get a polynomial ofdegree n−1. Repeatedly applying the Fundamental Theorem and Factor Theorem gives you n roots andn factors. Descartes’ Rule of SignsDescartes’ Rule of Signs can tell you how many positive and how many negative real zeroes thepolynomial has. This is a big labor‐saving device, especially when you’re deciding which possiblerational roots to pursue. To apply Descartes’ Rule of Signs, you need to understand the term variation in sign. When thepolynomial is arranged in standard form, a variation in sign occurs when the sign of a coef icient isdifferent from the sign of the preceding coef icient. (A zero coef icient is ignored.) For example, p(x) = x5 − 2x3 + 2x2 − 3x + 12has four variations in sign.Descartes’ Rule of Signs: The number of positive roots of p(x)=0 is either equal to the number of variations in sign of p(x), or less than that by an even number. The number of negative roots of p(x)=0 is either equal to the number of variations in sign of p(−x), or less than that by an even number.Example: Consider p(x) above. Since it has four variations in sign, there must be either four positiveroots, two positive roots, or no positive roots. 4 of 14
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Now form p(−x), by replacing x with (−x) in the above: p(−x) =(−x)5 − 2(−x)3 + 2(−x)2 − 3(−x) + 12 p(−x) = −x5 + 2x3 + 2x2 + 3x + 12p(−x) has one variation in sign, and therefore the original p(x) has one negative root. Since you knowthat p(x) must have a negative root, but it may or may not have any positive roots, you would look irstfor negative roots. p(x) is a ifth−degree polynomial, and therefore it must have ive zeros. Since x is not a factor, youknow that x=0 is not a zero of the polynomial. (For a polynomial with real coef icients, like this one,complex roots occur in pairs.) Therefore there are three possibilities: number of zeroes that are complex positive negative not real irst possibility 4 1 0 second possibility 2 1 2 third possibility 0 1 4 Complex RootsIf a polynomial has real coef icients, then either all roots are real or there are an even number ofnon-real complex roots, in conjugate pairs. For example, if 5+2i is a zero of a polynomial with real coef icients, then 5−2i must also be a zero ofthat polynomial. It is equally true that if (x−5−2i) is a factor then (x−5+2i) is also a factor. Why is this true? Because when you have a factor with an imaginary part and multiply it by itscomplex conjugate you get a real result: (x−5−2i)(x−5+2i) = x²−10x+25−4i² = x²−10x+29If (x−5−2i) was a factor but (x−5+2i) was not, then the polynomial would end up with imaginaries in itscoef icients, no matter what the other factors might be. If the polynomial has only real coef icients, thenany complex roots must occur in conjugate pairs. Irrational RootsFor similar reasons, if the polynomial has rational coef icients then the irrational roots involvingsquare roots occur (if at all) in conjugate pairs. If (x−2+√3) is a factor of a polynomial with rationalcoef icients, then (x−2−√3) must also be a factor. (To see why, remember how you rationalize abinomial denominator; or just check what happens when you multiply those two factors.) As Jeff Beckman pointed out (20 June 2006), this is emphatically not true for odd roots. Forinstance, x³−2 = 0 has three roots, 2^(1/3) and two complex roots. It’s an interesting problem whether irrationals involving even roots of order ≥4 must also occur in 5 of 14
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conjugate pairs. I don’t have an immediate answer. I’m working on a proof, as I have time. Multiple RootsWhen a given factor (x−r) occurs m times in a polynomial, r is called a multiple root or a root ofmultiplicity m If the multiplicity m is an even number, the graph touches the x axis at x=r but does not cross it. If the multiplicity m is an odd number, the graph crosses the x axis at x=r. If the multiplicity is 3, 5, 7, and so on, the graph is horizontal at the point where it crosses the axis.Examples: Compare these two polynomials and their graphs: f(x) = (x−1)(x−4)2 = x3 − 9x2 + 24x − 16 g(x) = (x−1)3(x−4)2 = x5 − 11x4 + 43x3 − 73x2 + 56x − 16These polynomials have the same zeroes, but the root 1 occurs with different multiplicities. Look at thegraphs:Both polynomials have zeroes at 1 and 4 only. f(x) has degree 3, which means three roots. You see fromthe factors that 1 is a root of multiplicity 1 and 4 is a root of multiplicity 2. Therefore the graph crossesthe axis at x=1 (but is not horizontal there) and touches at x=4 without crossing. By contrast, g(x) has degree 5. (g(x) = f(x) times (x‐1)2.) Of the ive roots, 1 occurs with multiplicity3: the graph crosses the axis at x=1 and is horizontal there; 4 occurs with multiplicity 2, and the graphtouches the axis at x=4 without crossing.Step 3. Quadratic FactorsWhen you have quadratic factors (Ax²+Bx+C), it may or may not be possible to factor them further. Sometimes you can just see the factors, as with x²−x−6 =(x+2)(x−3). Other times it’s not so obvious whether the quadraticcan be factored. That’s when the quadratic formula (shown atright) is your friend. 6 of 14
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For example, suppose you have a factor of 12x²−x−35. Can that be factored further? By trial and erroryou’d have to try a lot of combinations! Instead, use the fact that factors correspond to roots, andapply the formula to ind the roots of 12x²−x−35 = 0, like this: x = [ −(−1) ± √[1 − 4(12)(−35)] ] / 2(12) x = [ 1 ± √1681 ] / 24√1681 = 41, and therefore x = [ 1 ± 41 ] / 24 x = 42/24 or −40/24 x = 7/4 or −5/3If 7/4 and −5/3 are roots, then (x−7/4) and (x+5/3) are factors. Therefore 12x²−x−35 = (4x−7)(3x+5)What about x²−5x+7? This one looks like it’s prime, but how can you be sure? Again, apply the formula: x = [ −(−5) ± √[25 − 4(1)(7)] ] / 2(1) x = [ 5 ± √(−3) ] / 2What you do with that depends on the original problem. If it was to factor over the reals, then x²−5x+7is prime. But if that factor was part of an equation and you were supposed to ind all complex roots, youhave two of them: x = 5/2 + ((√3)/2)i, x = 5/2 − ((√3)/2)iSince the original equation had real coef icients, these complex roots occur in a conjugate pair.Step 4. Find One Factor or RootThis step is the heart of factoring a polynomial or solving a polynomial equation. There are a lot oftechniques that can help you to ind a factor. Sometimes you can ind factors by inspection (see the irst two sections that follow). This providesa great shortcut, so check for easy factors before starting more strenuous methods. Monomial FactorsAlways start by looking for any monomial factors you can see. For instance, if your function is f(x) = 4x6 + 12x5 + 12x4 + 4x3you should immediately factor it as f(x) = 4x3(x3 + 3x2 + 3x + 1)Getting the 4 out of there simpli ies the remaining numbers, the x3 gives you a root of x = 0 (withmultiplicity 3), and now you have only a cubic polynomial (degree 3) instead of a sextic (degree 6). Infact, you should now recognize that cubic as a special product, the perfect cube (x+1)3.When you factor out a common variable factor, be sure you remember it at the end when you’re listingthe factor or roots. x³+3x²+3x+1 = 0 has certain roots, but x³(x³+3x²+3x+1) = 0 has those same rootsand also a root at x=0. 7 of 14
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Special ProductsBe alert for applications of the Special Products. If you can apply them, your task becomes mucheasier. The Special Products are perfect square (2 forms): A² ± 2AB + B² = (A ± B)² sum of squares: A² + B² cannot be factored on the reals, in general (for exceptional cases see Factoring the Sum of Squares) difference of squares: A² − B² = (A + B)(A − B) perfect cube (2 forms): A³ ± 3A²B + 3AB² ± B³ = (A ± B)³ sum of cubes: A³ + B³ = (A + B)(A² − AB + B²) difference of cubes: A³ − B³ = (A − B)(A² + AB + B²)The expressions for the sum or difference of two cubes look as though they ought to factor further, butthey don’t. A²±AB+B² is prime over the reals.Consider p(x) = 27x³ − 64You should recognize this as p(x) = (3x)³ − 4³You know how to factor the difference of two cubes: p(x) = (3x−4)(9x²+12x+16)Bingo! As soon as you get down to a quadratic, you can apply the Quadratic Formula and you’re done.Here’s another example: q(x) = x6 + 16x3 + 64This is just a perfect square trinomial, but in x3 instead of x. You factor it exactly the same way: q(x) = (x3)2 + 2(8)(x3) + 82 q(x) = (x3 + 8)2And you can easily factor (x3+8)2 as (x+2)2(x2−2x+4)2. Rational RootsAssuming you’ve already factored out the easy monomial factors and special products, what do you doif you’ve still got a polynomial of degree 3 or higher? The answer is the Rational Root Test. It can show you some candidate roots when you don’t seehow to factor the polynomial, as follows. Consider a polynomial in standard form, written from highest degree to lowest and with onlyinteger coef icients: f(x) = anxn + ... + aoThe Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fractionp/q, where p is a factor of the trailing constant ao and q is a factor of the leading coef icient an.Example: p(x) = 2x4 − 11x3 − 6x2 + 64x + 32The factors of the leading coef icient (2) are 2 and 1. The factors of the constant term (32) are 1, 2, 4, 8, 8 of 14
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16, and 32. Therefore the possible rational zeroes are ±1, 2, 4, 8, 16, or 32 divided by 2 or 1: ±1/2, 1/1, 2/2, 2/1, 4/2, 4/1, 8/2, 8/1, 16/2, 16/1, 32/2, 32/1 reduced: ± ½, 1, 2, 4, 8, 16, 32What do we mean by saying this is a list of all the possible rational roots? We mean that no otherrational number, like ¼ or 32/7, can be a root of p(x) = 0.Caution: Don’t make the Rational Root Test out to be more than it is. It doesn’t say those rationalnumbers are roots, just that no other rational numbers can be roots. And it doesn’t tell you anythingabout whether some irrational or even complex roots exist. The Rational Root Test is only a startingpoint.Suppose you have a polynomial with non‐integer coef icients. Are you stuck? No, you can factor out theleast common denominator (LCD) and get a polynomial with integer coef icients that way. Example: (1/2)x³ − (3/2)x² + (2/3)x − 1/2The LCD is 1/6. Factoring out 1/6 gives the polynomial (1/6)(3x³ − 9x² + 4x − 3)The two forms are equivalent, and therefore they have the same roots. But you can’t apply the RationalRoot Test to the irst form, only to the second. The test tells you that the only possible rational roots are±1/3, 1, 3.Once you’ve identi ied the possible rational zeroes, how can you screen them? The brute‐force methodwould be to take each possible value and substitute it for x in the polynomial: if the result is zero thenthat number is a root. But there’s a better way. Use Synthetic Division to see if each candidate makes the polynomial equal zero. This is better forthree reasons. First, it’s computationally easier, because you don’t have to compute higher powers ofnumbers. Second, at the same time it tells you whether a given number is a root, it produces thereduced polynomial that you’ll use to ind the remaining roots. Finally, the results of synthetic divisionmay give you an upper or lower bound even if the number you’re testing turns out not to be a root. Sometimes Descartes’ Rule of Signs can help you screen the possible rational roots further. Forexample, the Rational Root Test tells you that if q(x) = 2x4 + 13x3 + 20x2 + 28x + 8has any rational roots, they must come from the list ±½, 1, 2, 4, 8. But don’t just start off substituting orsynthetic dividing. Since there are no sign changes, there are no positive roots. Are there any negativeroots? q(−x) = 2x4 − 13x3 + 20x2 − 28x + 8has four sign changes. Therefore there could be as many as four negative roots. (There could also betwo negative roots, or none.) There’s no guarantee that any of the roots are rational, but any root that isrational must come from the list −½, −1, −2, −4, −8. (If you have a graphing calculator, you can pre‐screen the rational roots by graphing thepolynomial and seeing where it seems to cross the x axis. But you still need to verify the rootalgebraically, to see that f(x) is exactly 0 there, not just nearly 0.)Remember, the Rational Root Test guarantees to ind all rational roots. But it will completely miss realroots that are not rational, like the roots of x²−2=0, which are ±√2, or the roots of x²+4=0, which are±2i.Finally, remember that the Rational Root Test works only if all 9 of 14
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coef icients are integers. Look again at this function, which isgraphed at right: p(x) = 2x4 − 11x3 − 6x2 + 64x + 32The Rational Root Theorem tells you that the only possiblerational zeroes are ±½, 1, 2, 4, 8, 16, 32. But suppose you factorout the 2 (as I once did in class), writing the equivalent function p(x) = 2(x4 − (11/2)x3 − 3x2 + 32x + 16)This function is the same as the earlier one, but you can nolonger apply the Rational Root Test because the coef icients are not integers. In fact −½ is a zero of p(x),but it did not show up when I (illegally) applied the Rational Root Test to the second form. My mistakewas forgetting that the Rational Root Theorem applies only when all coef icients of the polynomial areintegers. Graphical CluesBy graphing the function—either by hand or with a graphing calculator—you can get a sense of wherethe roots are, approximately, and how many real roots exist. Example: If the Rational Root Test tells you that ±2 are possible rational roots, you can look at thegraph to see if it crosses (or touches) the x axis at 2 or −2. If so, use synthetic division to verify that thesuspected root actually is a root. Yes, you always need to check—from the graph you can never be surewhether the intercept is at your possible rational root or just near it. Boundaries on RootsSome techniques don’t tell you the speci ic value of a root, but rather that a root exists between twovalues or that all roots are less than a certain number of greater than a certain number. This helpsnarrow down your search. Intermediate Value TheoremThis theorem tells you that if the graph of a polynomial is above the x axis for one value of x and belowthe x axis for another value of x, it must cross the x axis somewhere between. (If you can graph thefunction, the crossings will usually be be obvious.) Example: p(x) = 3x³ + 4x² − 20x −32The rational roots (if any) must come from the list ±1/3, 2/3, 1, 4/3, 2, 8/3, 4, 16/3, 8, 32/3, 16, 32.Naturally you’ll look at the integers irst, because the arithmetic is easier. Trying synthetic division, you ind p(1) = −45, p(2) = −22, and p(4) = 144. Since p(2) and p(4) have opposite signs, you know that thegraph crosses the axis between x=2 and x=4, so there is at least one root between those numbers. Inother words, either 8/3 is a root, or there root(s) between 2 and 4 are irrational. (In fact, syntheticdivision reveals that 8/3 is a root.)The Intermediate Value Theorem can tell you where there is a root, but it can’t tell you where there is 10 of 14
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no root. For example, consider q(x) = 4x² − 16x + 15q(1) and q(3) are both positive, but that doesn’t tell you whether the graph might touch or cross the axisbetween. (It actually crosses the axis twice, at x = 3/2 and x = 5/2.) Upper and Lower BoundsOne side effect of synthetic division is that even if the number you’re testing turns out as not a root, itmay tell you that all the roots are smaller or larger than that number: If you do synthetic division by a positive number a, and every number in the bottom row is positive or zero, then a is an upper bound for the roots, meaning that all the real roots are ≤ a. If you do synthetic division by a negative number b, and the numbers in the bottom row alternate sign, then b is a lower bound for the roots, meaning that all the real roots are ≥ b. What if the bottom row contains zeroes? A more complete statement is that alternating nonnegative and nonpositive signs, after synthetic division by a negative number, show a lower bound on the root. The next two examples clarify that.(By the way, the rule for lower bounds follows from the rule for upper bounds. Lower limits on roots ofp(x) equal upper limits on roots of p(−x), and dividing by (−x+r) is the same as dividing by −(x−r).)Example: q(x) = x3 + 2x2 − 3x − 4Using the Rational Root Test, you identify the only possible rational roots as ±4, ±2, and ±1. You decideto try −2 as a possible root, and you test it with synthetic division: -2 | 1 2 -3 -4 | -2 0 6 |------------------ 1 0 -3 2−2 is not a root of the equation f(x)=0. The third row shows alternating signs, and you were dividing bya negative number; however, that zero mucks things up. Recall that you have a lower bound only if thesigns in the bottom row alternate nonpositive and nonnegative. The 1 is positive (nonnegative), and the0 can count as nonpositive, but the −3 doesn’t qualify as nonnegative. The alternation is broken, and youdo not know whether there are roots smaller than −2. (In fact, graphical or numerical methods wouldshow a root around −2.5.) Therefore you need to try the lower possible rational root, −4: -4 | 1 2 -3 -4 | -4 8 -20 |------------------ 1 -2 5 -24Here the signs do alternate; therefore you know there are no roots below −4. (The remainder −24shows you that −4 itself isn’t a root.)Here’s another example: r(x) = x³ + 3x² − 3The Rational Root Test tells you that the possible rational roots are ±1 and ±3. With synthetic divisionfor −3: 11 of 14
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-3 | 1 3 0 -3 | -3 0 0 |------------------ 1 0 0 -3−3 is not a root, but the signs do alternate here, since the irst 0 counts as nonpositive and the second asnonnegative. Therefore −3 is a lower bound to the roots, meaning that the equation has no real rootslower than −3.Step 5. Divide by Your FactorRemember that r is a root if and only if x−r is a factor; this is the Factor Theorem. So if you want tocheck whether r is a root, you can divide the polynomial by x−r and see whether it comes out even(remainder of 0). Elizabeth Stapel has a nice example of dividing polynomials by long division. But it’s easier and faster to do synthetic division. If your synthetic division is a little rusty, youmight want to look at Dr. Math’s short Synthetic Division tutorial; if you need a longer tutorial,Elizabeth Stapel’s Synthetic Division is excellent. (Dr. Math also has a page on why Synthetic Divisionworks.) Synthetic division also has some side bene its. If your suspected root actually is a root, syntheticdivision gives you the reduced polynomial. And sometimes you also luck out and synthetic divisionshows you an upper or lower bound on the roots.You can use synthetic division when you’re dividing by a binomial of the form x−r for a constant r. Ifyou’re dividing by x−3, you’re testing whether 3 is a root and you synthetic divide by 3 (not −3). Ifyou’re dividing by x+11, you’re testing whether −11 is a root and you synthetic divide by −11 (not 11). Example: p(x) = 4x4 − 35x2 − 9You suspect that x−3 might be a factor, and you test it by synthetic division, like this: 3 | 4 0 -35 0 -9 | 12 36 3 9 |-------------------- 4 12 1 3 0Since the remainder is 0, you know that 3 is a root of p(x) = 0, and x−3 is a factor of p(x). But you knowmore. Since 3 is positive and the bottom row of the synthetic division is all positive or zero, you knowthat all the roots of p(x) = 0 must be ≤ 3. And you also know that p(x) = (x−3)(4x3 + 12x2 + x + 3)4x 3 + 12x2 + x + 3 is the reduced polynomial. All of its factors are also factors of the original p(x), butits degree is one lower, so it’s easier to work with.Step 6. Numerical MethodsWhen your equation has no more rational roots (or your polynomial has no more rational factors), you 12 of 14
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can turn to numerical methods to ind the approximate value of irrational roots: The Wikipedia article Root‐ inding algorithm has a decent summary, with pointers to speci ic methods. Many graphing calculators have a “Solve” or “Root” or “Zero” command to help you ind approximate roots. For instance, on the TI‐83 or TI‐84, you graph the function and then select [2nd] [Calc] [zero].Complete ExampleSolve for all complex roots: 4x³ + 15x − 36 = 0Step 1. The equation is already in standard form, with only zero on one side, and powers of x fromhighest to lowest. There are no common factors.Step 2. Since the equation has degree 3, there will be 3 roots. There is one variation in sign, and fromDescartes’ Rule of Signs you know there must be one positive root. Examine the polynomial with −xreplacing x: −4x³ − 15x − 36There are no variations in sign, which means there are no negative roots. The other two roots musttherefore be complex conjugates.Steps 3 and 4. The possible rational roots are unfortunately rather numerous: any of 1, 2, 3, 4, 6, 9, 12,18, 36 divided by any of 4, 2, 1. (Only positive roots are listed because you have already determined thatthere are no negative roots for this equation.) You decide to try 1 irst: 1 | 4 0 15 -36 | 4 4 19 |----------------- 4 4 19 -171 is not a root, so you test 2: 2 | 4 0 15 -36 | 8 16 62 |----------------- 4 8 31 26Alas, 2 is not a root either. But notice that f(1) = −17 and f(2) = 26. They have opposite signs, whichmeans that the graph crosses the x axis between x=1 and x=2, and a root is between 1 and 2. (In thiscase it’s the only root, since you have determined that there is one positive root and there are nonegative roots.) The only possible rational root between 1 and 2 is 3/2, and therefore either 3/2 is a root or theroot is irrational. You try 3/2 by synthetic division: 3/2 | 4 0 15 -36 | 6 9 36 |----------------- 4 6 24 0 13 of 14
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Hooray! 3/2 is a root. The reduced polynomial is 4x² + 6x + 24. In other words, (4x³ + 15x − 36) ÷ (x−3/2) = 4x² + 6x + 24The reduced polynomial has degree 2, so there is no need for more trial and error, and you continue tostep 5.Step 5. Now you must solve 4x² + 6x + 24 = 0First divide out the common factor of 2: 2x² + 3x + 12 = 0It’s no use trying to factor that quadratic, because you determined using Descartes’ Rule of Signs thatthere are no more real roots. So you use the quadratic formula: x = [ −3 ± √[9 − 4(2)(12)] ] / 2(2) x = [ −3 ± √(−87) ] / 4 x = −3/4 ± ((√87)/4)iStep 6. Remember that you found a root in an earlier step! The full list of roots is 3/2, −3/4 + ((√87)/4)i, −3/4 − ((√87)/4)iWhat’s New 8 Aug 2011: Add Dick Nickalls’ alternative solutions for cubic and quartic. (intervening changes suppressed) 15 Feb 2002: irst publicationthis page: http://oakroadsystems.com/math/polysol.htm 14 of 14
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