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Pre Calc Lesson 2.1

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

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Pre Calc Lesson 2.1

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  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 axn + bxn-1 + cxn-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 Degree Name Example 0 constant 5 1 Linear 3x + 2 2 Quadratic x2 + 3x – 6 3 Cubic x3 + 3x2 - 2x + 5 4 Quartic -3x4 + 7x3 + 10 5 Quintic x5 + 3x4 – 2x3 + 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: x2 – 4 x2 -- 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. • f(x) = 2x3 – 32x • Yes – a polynomial function. • to find the zeros: Let 0 = 2x3 – 32x • factor out a ‘2x’ 0 = 2x(x2 – 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) = 2x3 – 18x • x + 3 • 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. • Give the ‘zeros’ of ‘f’. • To find the zeros of anything that looks like this, set top = 0 • sooo 2x3 – 18x = 0 • just like before 2x(x2 – 9) = 0 • 2x(x – 3)(x + 3 = 0 • soooo x = 0, 3, - 3

  6. Example 3: • If P(x) = 3x4 – 7x3 – 5x2 + 9x + 10, find: • 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(81n4)-7(-27n3)-5(9n2)-27n+10 • 48 - 56 – 20 + 18 + 10 | 243n4 + 189n3 – 45n2 – 27n + 10 • - 8 - 2 + 10 • -10 + 10 • 0 • Therefore: • P(2) = 0 & P(-3n) = 243n4 + 189n3 – 45n2- 27n + 10

  7. Example 4: Use a process called ‘synthetic substitution’ • to evaluate the following: • a) If P(x) = 3x4 – 7x3 – 5x2 + 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) • If S(x) = 3x4 – 5x2 + 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

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