3.2 The Factor Theorem and The Remainder Theorem

1 The Factor Theorem and The Remainder The Factor Theorem and The Remainder TheoremSuppose we wish to find the zeros off(x) =x3+ 4x2 5x 14. Settingf(x) = 0 results in thepolynomial equationx3+ 4x2 5x 14 = 0. Despite all of the factoring techniques we learned1in Intermediate Algebra, this equation foils2us at every turn. If we graphfusing the graphingcalculator, we getThe graph suggests that the function has three zeros, one of which isx= 2. It s easy to showthatf(2) = 0, but the other two zeros seem to be less friendly. Even though we could use the Zero command to find decimal approximations for these, we seek a method to find the remainingzerosexactly.

2 Based on our experience, ifx= 2 is a zero, it seems that there should be a factorof (x 2) lurking around in the factorization off(x). In other words, we should expect thatx3+ 4x2 5x 14 = (x 2)q(x), whereq(x) is some other polynomial. How could we find suchaq(x), if it even exists? The answer comes from our old friend, polynomial division. Dividingx3+ 4x2 5x 14 byx 2 givesx2+ 6x+ 7x 2x3+ 4x2 5x 14 (x3 2x2)6x2 5x (6x2 12x)7x 14 (7x 14)0As you may recall, this meansx3+ 4x2 5x 14 = (x 2)(x2+ 6x+ 7), so to find the zeros off,we now solve (x 2)(x2+ 6x+ 7)= 0. We getx 2 = 0 (which gives us our known zero,x= 2)as well asx2+ 6x+ 7 = 0.

3 The latter doesn t Factor nicely, so we apply the Quadratic Formula togetx= 3 2. The point of this section is to generalize the technique applied here. First up isa friendly reminder of what we can expect when we divide probably forgot2pun intended258 Polynomial FunctionsTheorem Polynomial Division:Supposed(x) andp(x) are nonzero polynomials wherethe degree ofpis greater than or equal to the degree ofd. There exist two unique polynomials,q(x) andr(x), such thatp(x) =d(x)q(x) +r(x),where eitherr(x) = 0 or the degree ofrisstrictly less than the degree you may recall, all of the polynomials in Theorem have special names.

4 The polynomialpis called thedividend;dis thedivisor;qis thequotient;ris theremainder. Ifr(x) = 0 thendis called afactorofp. The proof of Theorem is usually relegated to a course in AbstractAlgebra,3but we can still use the result to establish two important facts which are the basis of therest of the The Remainder Theorem :Supposepis a polynomial of degree at least 1andcis a real number. Whenp(x) is divided byx cthe Remainder isp(c).The proof of Theorem is a direct consequence of Theorem When a polynomial is dividedbyx c, the Remainder is either 0 or has degree less than the degree ofx c.

5 Sincex cis degree1, the degree of the Remainder must be 0, which means the Remainder is a constant. Hence, ineither case,p(x) = (x c)q(x) +r, wherer, the Remainder , is a real number, possibly 0. It followsthatp(c) = (c c)q(c) +r= 0 q(c) +r=r, so we getr=p(c) as required. There is one last lowhanging fruit 4to collect which we present The Factor Theorem :Supposepis a nonzero polynomial. The real numbercis a zero ofpif and only if (x c) is a Factor ofp(x).The proof of The Factor Theorem is a consequence of what we already know. If (x c) is a factorofp(x), this meansp(x) = (x c)q(x) for some polynomialq.

6 Hence,p(c) = (c c)q(c) = 0, socis a zero ofp. Conversely, ifcis a zero ofp, thenp(c) = 0. In this case, The Remainder Theoremtells us the Remainder whenp(x) is divided by (x c), namelyp(c), is 0, which means (x c) is afactor ofp. What we have established is the fundamental connection between zeros of polynomialsand factors of the things The Factor Theorem tells us, the most pragmatic is that we had better find a moreefficient way to divide polynomials by quantities of the formx c. Fortunately, people like Ruffiniand Hornerhave already blazed this trail. Let s take a closer look at the long division we performedat the beginning of the section and try to streamline it.

7 First off, let s change all of the subtractionsinto additions by distributing through the , Virginia, there are Algebra courses more abstract than this hates this expression and Carl included it just to annoy The Factor Theorem and The Remainder Theorem259x2+ 6x+ 7x 2x3+ 4x2 5x 14 x3+ 2x26x2 5x 6x2+ 12x7x 14 7x+140 Next, observe that the terms x3, 6x2and 7xare the exact opposite of the terms above algorithm we use ensures this is always the case, so we can omit them without losing anyinformation. Also note that the terms we bring down (namely the 5xand 14) aren t reallynecessary to recopy, so we omit them, + 6x+ 7x 2x3+4x2 5x 142x26x212x7x140 Now, let s move things up a bit and, for reasons which will become clear in a moment, copy thex3into the last + 6x+ 7x 2x3+4x2 5x 142x212x14x36x27x0 Note that by arranging things in this manner, each term in the last row is obtained by adding thetwo terms above it.

8 Notice also that the quotient polynomial can be obtained by dividing each ofthe first three terms in the last row byxand adding the results. If you take the time to work backthrough the original division problem, you will find that this is exactly the way we determined thequotient polynomial. This means that we no longer need to write the quotient polynomial down,nor thexin the divisor, to determine our answer. 2x3+4x2 5x 142x212x14x36x27x0260 Polynomial FunctionsWe ve streamlined things quite a bit so far, but we can still do more. Let s take a moment toremind ourselves where the 2x2, 12xand 14 came from in the second row.

9 Each of these terms wasobtained by multiplying the terms in the quotient,x2, 6xand 7, respectively, by the 2 inx 2,then by 1 when we changed the subtraction to addition. Multiplying by 2 then by 1 is thesame as multiplying by 2, so we replace the 2 in the divisor by 2. Furthermore, the coefficients ofthe quotient polynomial match the coefficients of the first three terms in the last row, so we nowtake the plunge and write only the coefficients of the terms to get21 4 5 14212141 670We have constructed asynthetic division tableaufor this polynomial division problem. Let s re-work our division problem using this tableau to see how it greatly streamlines the division dividex3+ 4x2 5x 14 byx 2, we write 2 in the place of the divisor and the coefficients ofx3+ 4x2 5x 14 in for the dividend.

10 Then bring down the first coefficient of the 4 5 1421 4 5 14 1 Next, take the 2 from the divisor and multiply by the 1 that was brought down to get 2. Writethis underneath the 4, then add to get 4 5 14 2121 4 5 14 21 6 Now take the 2 from the divisor times the 6 to get 12, and add it to the 5 to get 4 5 14 2121 621 4 5 14 2121 67 Finally, take the 2 in the divisor times the 7 to get 14, and add it to the 14 to get 4 5 14 212141 6721 4 5 14 212141 The Factor Theorem and The Remainder Theorem261 The first three numbers in the last row of our tableau are the coefficients of the quotient , we started with a third degree polynomial and divided by a first degree polynomial.