VIEWS: 36 PAGES: 12 POSTED ON: 9/29/2012
1.5 - Factoring Polynomials - The Factor Theorem MCB4U - Santowski (A) Review • when a number divides evenly into another number, then it is a factor of the number that it has divided • ex. since 4 divides evenly into 12, 4 is said to be factor of 12 and since 5 does not divide evenly into 12, 5 is said not to be a factor of 12. • if this concept holds true for numbers, then the concept should also hold true for polynomials • Divide x3 – x2 - 14x + 24 by x - 2 and notice what the remainder is? Then evaluate P(2). What must be true about (x - 2)? • Now divide x3 – x2 - 14x + 24 by x + 3 and notice what the remainder is? Then evaluate P(-3). What must be true about (x+3)? • Now graph f(x) = x3 – x2 - 14x + 24 and see what happens at x = 2 and x = -3 • So our conclusion is that x - 2 is a factor of x3 – x2 - 14x + 24, whereas x + 3 is not a factor of x3 – x2 - 14x + 24 (A) Review – Graph of P(x) = x3 – x2 - 14x + 24 (B) The Factor Theorem • We can use the ideas developed in the review to help us to draw a connection between the polynomial, its factors, and its roots. • What we have seen in our review are the key ideas of the Factor Theorem - in that if we know a root of an equation, we know a factor and the converse, that if we know a factor, we know a root. • The Factor Theorem is stated as follows: x - a is a factor of f(x) if and only if f(a) = 0. To expand upon this idea, we can add that ax - b is a factor of f(x) if and only if f(b/a) = 0. • Working with polynomials, (x + 1) is a factor of x2 + 2x + 1 because when you divide x2 + 2x + 1 by x + 1 you get a 0 remainder and when you substitute x = -1 into x2 + 2x + 1 , you get 0 (C) Examples • ex 1. Show that x - 2 is a factor of x3 - 7x + 6 • ex. 2. Show that -2 is a root of 2x3 + x2 - 2x + 8 = 0. Find the other roots of the equation. (Show with GC) • ex. 3. Factor x3 + 1 completely • ex. 4. Is x - 2 a factor of x4 – 5x2 + 6? (D) Rational Roots of Polynomial Equations • Our previous examples were slightly misleading … as in too easy • In each example, you were given a root to test …. so what would you do if no root was given for you to test and you had to suggest a root to test in the first place?? • Consider this example with quadratics 6x2 + 7x – 3 which when factored becomes (2x+3)(3x-1) so the roots would be –3/2 and 1/3 • Make the following observation that the numerator of the roots (-3,1) are factors of the constant term (-3) while the denominator of the roots (2,3) are factors of the leading coefficient (6) • We can test this idea with other polynomials we will find the same pattern that the roots are in fact some combination of the factors of the leading coefficient and the constant term (E) Rational Root Theorem • Our previous observation (although limited in development) leads to the following theorem: • Given that P(x) = anxn + an-1xn-1 + ….. + a1x1 + a0, if P(x) = 0 has a rational root of the form a/b and a/b is in lowest terms, then a must be a divisor of a0 and b must be a divisor of an (E) Rational Root Theorem • So what does this theorem mean? • If we want to factor the polynomial P(x) = 2x3 – 5x2 + 22x – 10, then we first need to find a value a/b such that P(a/b) = 0 • So the factors of the leading coefficient are {+1,+2} which are then the possible values for a • The factors of the constant term, -10, are {+1,+2,+5,+10} which are then the possible values for b • Thus the possible ratios a/b which we can test using the Factor Theorem are {+1,+½ ,+2,+5/2,+5,+10} • As it then turns out, P(½) turns out to give P(x) = 0, meaning that x – ½ (or 2x – 1) is a factor of P(x) • From this point on, we can then do the synthetic division (using ½) to find the quotient and then possibly other factor(s) of P(x) (F) Further Examples • Ex 1 Factor P(x) = 2x3 – 9x2 + 7x + 6 to find the roots of P(x) • Ex 2 Factor P(x) = 3x3 – 7x2 + 8x – 2 = 0 to determine the roots if (i) x R and if (ii) x C • ex 3 Graph f(x) = 3x3 + x2 - 22x - 24 using intercepts, points, and end behaviour. Approximate turning points, max/min points, and intervals of increase and decrease. (G) Internet Links • The Factor Theorem from the Math Page • College Algebra Tutorial on The Factor Theorem • The Factor Theorem from Purple Math (H) Factoring By Grouping • We can sometimes simplify the factoring of higher order polynomials if we can group terms within the polynomial to facilitate easier factoring of the polynomial • ex. Factor x4 - 6x3 + 2x2 - 12x by grouping • We will group it as: • = (x4 - 6x3) + (2x2 - 12x) • = x3(x - 6) + 2x(x – 6) • = (x - 6)(x3 + 2x) • = (x - 6)(x)(x2 + 2) (I) Homework • Nelson text, page 50; Q2,3,6,8,9 (all eol) and 10-13,14,16,18