Remainder and Factor theorems

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M4L4 The factor theorem and Remainder theorem:

M4L4 The factor theorem and Remainder theorem Adapted from Reynaldo B Pantino’s presentation

Objectives::

Objectives: To identify whether a given linear binomial is a factor of a polynomial function. To determine the remainder if you divide by a linear binomial.

Slide3:

Is 3 is a factor of 27? If yes, what makes it a factor of 27? It divides evenly with no remainder.

Slide4:

The Remainder Theorem states that when the polynomial P(x) is divided by x – c, the remainder is P(c). Example: When P(x) = x 3 – x 2 – 4x + 4 is divided by x – 2, the remainder is 0. That is, P(2) = 0.

Slide5:

The Remainder Theorem states that when the polynomial P(x) is divided by x – c, the remainder is P(c). Example: When P(x) = x 3 – x 2 – 4x + 4 is divided by x – 2, we have; 2 1 -1 -4 4 1 1 -2 0 2 2 -4 remainder

Slide6:

Notice that P(c) = 0, using synthetic division P(x) = (x – c) ● Q(x) + R becomes P(x) = (x – c) ● Q(x) + 0 P(x) = (x – c) ● Q(x). 2 1 -1 -4 4 1 1 -2 0 2 2 -4 remainder

Slide7:

FACTOR THEOREM Let P(x) be a polynomial. If P(c) = 0, where c is a real number, then (x – c) is a factor of P(x). Conversely, if (x – c) is a factor of P(x), then P(c) = 0. Since the theorem has a converse, the proof consists of two parts. a.) If (x – c) is a factor of P(x), then P(c) = 0. b.) If P(c) = 0, then (x – c) is a factor of P(x).

Examples::

Examples: 1. Show that x + 1 is a factor of 2x 3 + 5x 2 – 3. Solution: Let P(x) = 2x 3 + 5x 2 – 3 P(-1) = 2(-1) 3 + 5(-1) 2 – 3 P(-1) = -2 + 5 – 3 P(-1) = 0 By the Remainder Theorem, we know there is no remainder because P(-1) = 0. By Factor theorem, x + 1 is a factor of 2x 3 + 5x 2 – 3 .

Examples::

2. Check is x + 2 is a factor of x 4 + x 3 – x 2 – x - 18. Solution: Let P(x) = x 4 + x 3 – x 2 – x - 18 P(-2) = (-2) 4 + (-2) 3 – (-2) 2 – (-2) – 18 P(-2) = 16 - 8 – 4 + 2 – 18 P(-2) = -12 By the Remainder Theorem, we know the remainder is -12 because P(-2)=-12. By Factor theorem, x – 2 is a factor of x 4 + x 3 – x 2 – x - 18 . Examples:

Examples::

3. Show that f(x) = x 3 + x 2 – 5x + 3 is divisible by x + 3. Solution : Let, f(-3) = (-3) 3 + (-3) 2 – 5 (-3) + 3 = 0 = -27 + 9 + 15 + 3 = -27 +27 = 0 By the Remainder Theorem, we know there is no remainder because f(-3) = 0. By Factor theorem, x + 3 is a factor of x 3 + x 2 – 5x + 3 meaning it is divisible by x+3 . Examples:

PRACTICE EXERCISES:

PRACTICE EXERCISES  

ANSWERS: :

ANSWERS: 1. Show that x 3 –6x 2 +11x-6 is divisible by x-1   Solution:   X=1   = (1) 3 -6(1) 2 +11(1)-6 = 1-6+11-6 = -5+5 = 0 .      2. x 6  + 5 x 5  + 5 x 4  + 5 x 3  + 2 x 2  – 10 x  – 8; x= -4   Solution:   =  (-4) 6  + 5 (-4) 5  + 5 (-4) 4  + 5 (-4) 3  + 2 (-4) 2  – 10 (-4)  – 8 = 4096-5120+1280-320+32+40-8 =-1024+960+72-8 = -64+64 = 0.  

Slide13:

3. x 4 +7x 3 +3x 2 -63x-108; x=-3   Solution:   = (-3) 4 +7(-3) 3 +3(-3) 2 -63(-3)-108 = 81-189+27+189-108 = -108+216-108 = 108-108 = 0.   4. X 3 +4x 2 -9x-36; x=3   Solution: = (3) 3 +4(3) 2 -9(3)-36 = 27+36-27-36 = 63-63 = 0.   5. X 3 +4x 2 -9x-36; x=1   Solution:   = (1) 3 +4(1) 2 -9(1)-36 = 1+4-9-36 = 5-45 = 40. X-1 does not divide X 3 +4x 2 -9x-36.