1. Pre Calc Lesson 2.1 Zeros and factors of Polynomial functions A Polynomial in ‘x’ is any expression that can be written In the form of ax n + bx n-1 + cx n-2 + … + jx + k ----- ‘n’ must be a non-negative number a,b,c,… are called the coefficients of the polynomial Leading term -- the term containing the highest power of ‘x’ The coefficient of the leading term -- Leading coefficient Degree -- The power on ‘x’ contained in the leading term

2. Polynomials of the first few degrees have special names DegreeNameExample 0 constant 5 1 Linear 3x + 2 2 Quadratic x 2 + 3x – 6 3 Cubic x 3 + 3x 2 - 2x + 5 4 Quartic -3x 4 + 7x 3 + 10 5 Quintic x 5 + 3x 4 – 2x 3 + 11 To identify a particular term in a polynomial, we use the name associated with the power on ‘x’ that identifies that particular term. Example: x 2 – 4 x 2 -- called the ‘quadratic’ term - 4 -- called the ‘constant’ term There is no linear term.

3. ** Every polynomial P define a function!** Any value of ‘x’ for which P(x) = 0 is called a ‘root’ of the equation and a ‘zero’ of the function. All situations we have encountered so far are classified as ‘polynomial functions’. There are two situations which cause a polynomial not to be a function 1. A variable in the denominator of a fraction: P(x) = 3 x 2. A variable under the radical:

4. Example 1: State whether each function is a polynomial function. Give the zeros of the function, if they exist. a)f(x) = 2x 3 – 32x Yes – a polynomial function. to find the zeros: Let 0 = 2x 3 – 32x factor out a ‘2x’ 0 = 2x(x 2 – 16) continue factoring 0 = 2x(x – 4)(x + 4) zeros: x = 0, 4, - 4 b) g(x) = x + 1 x – 1 Not a Polynomial function (a variable in denominator) But there are still ‘zeros’ to this function. a fraction = 0 when the ‘top’ = 0 soooo – set x + 1 = 0 zero  x = - 1

5. Example 2: Let f(x) = 2x 3 – 18x x + 3 a)Tell whether ‘f’ is a polynomial function. Not a polynomial function -- a variable is in the denominator! b) List each value for ‘x’ for which ‘f’ is undefined to determine ‘undefined’ set ‘bottom’ ≠ 0 sooo x + 3 ≠ 0 x ≠ - 3 sooo if x = 3, this causes the ‘f’ to be undefined. c)Give the ‘zeros’ of ‘f’. To find the zeros of anything that looks like this, set top = 0 sooo 2x 3 – 18x = 0 just like before 2x(x 2 – 9) = 0 2x(x – 3)(x + 3 = 0 soooo x = 0, 3, - 3

6. Example 3: If P(x) = 3x 4 – 7x 3 – 5x 2 + 9x + 10, find: a)P(2) b) P(-3n) both of these are just ‘plug and chug’ problems. 3(2) 4 -7(2) 3 -5(2) 2 +9(2)+10 | (-3n) 4 -7(-3n) 3 -5(-3n) 2 +9(-3n)+10 3(16)-7(8)-5(4)+18+10 | 3(81n 4 )-7(-27n 3 )-5(9n 2 )-27n+10 48 - 56 – 20 + 18 + 10 | 243n 4 + 189n 3 – 45n 2 – 27n + 10 - 8 - 2 + 10 -10 + 10 0 Therefore: P(2) = 0 & P(-3n) = 243n 4 + 189n 3 – 45n 2 - 27n + 10

7. Example 4: Use a process called ‘synthetic substitution’ to evaluate the following: a) If P(x) = 3x 4 – 7x 3 – 5x 2 + 9x + 10, find P(2) 3 - 7 - 5 + 9 + 10 6 - 2 - 14 - 10 2) 3 - 1 - 7 - 5 0 P(2) = 0  (2,0) b)If S(x) = 3x 4 – 5x 2 + 9x + 10, find S(- 2) 3 0 - 5 + 9 + 10 -6 12 -14 10 - 2) 3 -6 7 -5 20 S(-2) = 20  (- 2,20) Hw : pg 56 #1-31 odd