Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

GI nternet Links The Factor Theorem from the Math Page College Algebra Tutorial on The Factor Theorem The Factor Theorem from Purple Math HF actoring By Grouping We by VII9jovw

VIEWS: 36 PAGES: 12

									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

								
To top