Unit 9: Polynomial Operations Synthetic division notes – day 1

Yesterday… Re-visited solving quadratics Explored the relationship between the degree of a polynomial and the number of roots

If you were asked to solve: 3x2 + x – 2 = 0 x3 + x2 – 12x = 0

What about this one? Hmmmm…..

Synthetic Division Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

Is it a factor? Sometimes, we are just trying to determine IF a linear binomial is a factor We can use Synthetic division to determine this by looking at the remainder! If the remainder is 0, then it is a factor If the remainder is not zero, then it is not a factor

Example 1: Determining Whether a Linear Binomial is a Factor Determine whether the given binomial is a factor of the polynomial P(x). A. (x + 1); (x2 – 3x + 1) B. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–1) by synthetic substitution. Find P(–2) by synthetic substitution. –1 1 –3 1 –1 4 –2 3 6 0 –5 –10 1 –4 5 –6 10 3 –5 P(–1) = 5 P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

Factor to find all the roots If we are being asked to factor a higher order polynomial, we will use a combination of synthetic division and factoring quadratics to get our factors.

Example: Use division to factor

You Try: Use division to factor