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1.5 - Factoring Polynomials - The Factor Theorem

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

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1.5 - Factoring Polynomials - The Factor Theorem

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  1. 1.5 - Factoring Polynomials - The Factor Theorem MCB4U - Santowski

  2. (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

  3. (A) Review – Graph of P(x) = x3 – x2 - 14x + 24

  4. (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 thatax - 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

  5. (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?

  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

  7. (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

  8. (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 ratiosa/bwhich 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)

  9. (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.

  10. (G) Internet Links • The Factor Theorem from the Math Page • College Algebra Tutorial on The Factor Theorem • The Factor Theorem from Purple Math

  11. (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) 

  12. (I) Homework • Nelson text, page 50; Q2,3,6,8,9 (all eol) and 10-13,14,16,18

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