7.1 Synthetic Division & the Remainder & Factor Theorems

Nested Form: There is a way to express a polynomial in a different way by factoring out what we can at different places in an equation Ex: We can write Now, let’s try this one: Ex 1) Write polynomial in nested form: (Hint, start from right side & work your way in)

Synthetic Division/Substitution: A process that takes up less space than long division & uses only the coefficients of each polynomial **Must write in descending order & fill in any missing terms with 0 ** This process can tell us: (1) if a polynomial is a factor (2) if a number is a root (3) what the value of an expression is Ex 2) Evaluate (answer is the number in the last spot)  Remainder Theorem –3 4 3 0 1 –7 9 ↓ –12 27 –81 240 –699 –3 x 4 = –12 4 –9 27 –80 233 –690 f (–3) = –690

Ex 3) Divide using synthetic division & then express the polynomial as a product of a divisor & a quotient with a remainder. * Note: dividing by x – 1 makes x – 1 = 0  x = 1 and 1 goes in the box 1 6 –2 5 1 ↓ 6 4 9 6 4 9 10 Coeff of Quotient Remainder Powers of x have decreased by 1 (6x2 + 4x + 9)(x – 1) + 10

Stemming from this example … if the result was a divisor & quotient being multiplied and there was nothing added or subtracted, the last number would be a zero! What does this mean?! It means if we do synthetic division / substitution and the end result is 0, we have a factor / root  Factor Theorem Ex 4) Determine whether x – 4 is a factor of 4 1 –4 3 –3 –36 ↓ 4 12 36 1 3 9  YAY!!  YES

1 1 3 –1 –3 ↓ 1 4 3 1 4 3 Yes, so (x – 1) is one factor We can factor a polynomial completely by first using synthetic to find 1 factor & then deal with the depressed equation. Ex 5) Express as a product of linear factors Try 1 1 1 3 –1 –3 ↓ 1 4 3 1 4 3 Yes, so (x – 1) is one factor Now use x2 + 4x + 3 and factor it! (x + 3)(x + 1) g(x) = (x – 1)(x + 3)(x + 1)

8 –16 16 –27 18 ↓ 12 –6 15 –18 8 –4 10 –12 Dealing with fractions Ex 6) Divide Factor is 2x – 3 Root is 2x – 3 = 0 8 –16 16 –27 18 ↓ 12 –6 15 –18 8 –4 10 –12 Factor out 2 So, answer is:

2 2 5 k –16 ↓ 4 18 36+2k 2 9 18+k 20+2k Want this to be 0, so Solving for an unknown inside an equation Ex 7) Determine k so that has x – 2 as a factor x – 2 = 0 x = 2 2 2 5 k –16 ↓ 4 18 36+2k 2 9 18+k 20+2k Want this to be 0, so 20 + 2k = 0 20 = –2k –10 = k

Summary (with proper vocab): If in polynomial P(x), P(a) = 0, then x = a is a zero of P(x) x = a is a solution of the equation P(x) = 0 x – a is a factor of P(x)

Homework #701 Pg 334 #3–15 odd, 16, 23, 27, 31–37 odd, 47, 48, 51, 55, 57, 59, 61