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 x 3 – x 2 - 14x + 24 by x - 2 and notice what the remainder is?  Then evaluate P(2). What must be true about (x - 2)? Now divide x 3 – x 2 - 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) = x 3 – x 2 - 14x + 24 and see what happens at x = 2 and x = -3 So our conclusion is that x - 2 is a factor of x 3 – x 2 - 14x + 24, whereas x + 3 is not a factor of x 3 – x 2 - 14x + 24

3. (A) Review – Graph of P(x) = x 3 – x 2 - 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.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.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.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 x 2 + 2x + 1 because when you divide x 2 + 2x + 1 by x + 1 you get a 0 remainder and when you substitute x = -1 into x 2 + 2x + 1, you get 0Working with polynomials, (x + 1) is a factor of x 2 + 2x + 1 because when you divide x 2 + 2x + 1 by x + 1 you get a 0 remainder and when you substitute x = -1 into x 2 + 2x + 1, you get 0

5. (C) Examples ex 1. Show that x - 2 is a factor of x 3 - 7x + 6ex 1. Show that x - 2 is a factor of x 3 - 7x + 6 ex. 2. Show that -2 is a root of 2x 3 + x 2 - 2x + 8 = 0. Find the other roots of the equation. (Show with GC)ex. 2. Show that -2 is a root of 2x 3 + x 2 - 2x + 8 = 0. Find the other roots of the equation. (Show with GC) ex. 3. Factor x 3 + 1 completelyex. 3. Factor x 3 + 1 completely ex. 4. Is x -  2 a factor of x 4 – 5x 2 + 6?ex. 4. Is x -  2 a factor of x 4 – 5x 2 + 6?

6. (D) Rational Roots of Polynomial Equations Our previous examples were slightly misleading … as in too easyOur 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??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  6x 2 + 7x – 3 which when factored becomes (2x+3)(3x-1) so the roots would be –3/2 and 1/3Consider this example with quadratics  6x 2 + 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)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 termWe 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:Our previous observation (although limited in development) leads to the following theorem: Given that P(x) = a n x n + a n-1 x n-1 + ….. + a 1 x 1 + a 0, 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 a 0 and b must be a divisor of a nGiven that P(x) = a n x n + a n-1 x n-1 + ….. + a 1 x 1 + a 0, 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 a 0 and b must be a divisor of a n

8. (E) Rational Root Theorem So what does this theorem mean?So what does this theorem mean? If we want to factor the polynomial P(x) = 2x 3 – 5x 2 + 22x – 10, then we first need to find a value a/b such that P(a/b) = 0If we want to factor the polynomial P(x) = 2x 3 – 5x 2 + 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 aSo 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 bThe 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}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)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)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) = 2x 3 – 9x 2 + 7x + 6 to find the roots of P(x)Ex 1  Factor P(x) = 2x 3 – 9x 2 + 7x + 6 to find the roots of P(x) Ex 2  Factor P(x) = 3x 3 – 7x 2 + 8x – 2 = 0 to determine the roots if (i) x  R and if (ii) x  CEx 2  Factor P(x) = 3x 3 – 7x 2 + 8x – 2 = 0 to determine the roots if (i) x  R and if (ii) x  C ex 3  Graph f(x) = 3x 3 + x 2 - 22x - 24 using intercepts, points, and end behaviour. Approximate turning points, max/min points, and intervals of increase and decrease.ex 3  Graph f(x) = 3x 3 + x 2 - 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 PageThe Factor Theorem from the Math PageThe Factor Theorem from the Math PageThe Factor Theorem from the Math Page College Algebra Tutorial on The Factor TheoremCollege Algebra Tutorial on The Factor TheoremCollege Algebra Tutorial on The Factor TheoremCollege Algebra Tutorial on The Factor Theorem The Factor Theorem from Purple MathThe Factor Theorem from Purple MathThe Factor Theorem from Purple MathThe 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 polynomialWe 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 x 4 - 6x 3 + 2x 2 - 12x by groupingex. Factor x 4 - 6x 3 + 2x 2 - 12x by grouping We will group it as:We will group it as: = (x 4 - 6x 3 ) + (2x 2 - 12x)= (x 4 - 6x 3 ) + (2x 2 - 12x) = x 3 (x - 6) + 2x(x – 6)= x 3 (x - 6) + 2x(x – 6) = (x - 6)(x 3 + 2x)= (x - 6)(x 3 + 2x) = (x - 6)(x)(x 2 + 2)= (x - 6)(x)(x 2 + 2)

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