# 6.5 The Remainder and Factor Theorems

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6.5 The Remainder and Factor Theorems
OBJ: use long division to divide polynomials Do Now: Use long division to divide. Round to the nearest hundredth. 1) divided by 36 2) 12 divided by 5

Polynomial Long Division
If a term is missing in standard form, fill it in with a 0 coefficient. Ex 1: 2x4 + 3x3 + 5x – 1 x2 – 2x + 2 2x2 2x4 = 2x2 x2

-( ) 2x2 +7x +10 2x4 -4x3 +4x2 7x3 - 4x2 +5x -( ) 7x3 - 14x2 +14x
-( ) 2x4 -4x3 +4x2 7x3 - 4x2 +5x -( ) 7x x2 +14x 10x2 - 9x -1 7x3 = 7x x2 -( ) 10x2 - 20x +20 11x - 21 remainder

The answer is written: Quotient + Remainder over Divisor
2x2 + 7x x – x2 – 2x + 2 Ex 2: y4 + 2y2 – y y2 – y + 1 y2 + y y2 – y + 1

Remainder Theorem: If a polynomial is divisible by (x – k), then the remainder is f(k) ONLY when you divide by x-k, check if the remainder is f(k) by directly substituting k back into the eqn!!! On your homework, this is possible on problems #6 and # 7 HW:

Divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2
continued… Do Now: Use long division to divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2 3x2 + 4x x – 2 Synthetic Division (like synthetic substitution) Ex 3: Divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2

f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2
6 8 20 3 4 10 R: 15 + 15 x-2 3x2 + 4x + 10

Divide x3 + 2x2 – 6x -9 by x-2 1 2 -6 -9 2 Ex 4: 2 8 4 1 4 2 R:-5
2 8 4 1 4 2 R:-5 x2 + 4x x-2

Divide x3 + 2x2 – 6x -9 by x+3 1 2 -6 -9 -3 Ex 5: -3 3 9 1 -1 -3 R: 0
-3 -3 3 9 1 -1 -3 R: 0 x2 – x - 3

Factor Theorem: A polynomial has the factor x-k if f(k)=0
Factor Theorem: A polynomial has the factor x-k if f(k)=0 ** if the remainder is zero, x-k is a factor since it divides evenly ** it also means that k is a “zero” (a solution) of the function! HW:

Factoring a polynomial
continued… Factoring a polynomial Do Now: Use synthetic division to divide f(x)= x4 -6x3 -40x +33 by x - 7 x3+ x2 + 7x x – 7 Factor f(x) = 2x3 + 11x2 + 18x + 9 Given f(-3)=0 *Since the remainder is zero, x – k is a factor! x-(-3) or x+3 *So use synthetic division to find the others!! Ex 6:

-3 -15 -9 -6 2 5 3 R: 0 (x + 3)(2x2 + 5x + 3) *keep factoring (bustin ‘da ‘b’) gives you: (x+3)(2x+3)(x+1)

Factor f(x)= 3x3 + 13x2 + 2x -8 Given f(-4)=0 (x + 1)(3x – 2)(x + 4)
Ex 7: Factor f(x)= 3x3 + 13x2 + 2x -8 Given f(-4)=0 (x + 1)(3x – 2)(x + 4)

Finding the zeros of a polynomial function
Ex 8: One zero of f(x) = x3 – 2x2 – 9x +18 is x=2. Find the others. *if x=2 is a zero, then x-2 is a factor *Use synthetic division to factor completely. *then solve (x-2)(x2-9) (x-2)(x+3)(x-3) x=2 x=-3 x=3

One zero of f(x) = x3 + 6x2 + 3x -10 is x=-5. Find the others.
Ex 9: One zero of f(x) = x3 + 6x2 + 3x -10 is x=-5. Find the others. (x-2)(x+1)(x+5) x=2 x=-1 x=-5 HW: