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Section R4 B Zeros of a Function Definition: Zero (sometimes called a Root) of a Function are the values in a function's domain where the function value equals zero. If you visualize a function like: the places where I cross the x-axis x = {x 1 , x2 , x3 } are the zeros. When we work with polynomials, the number of zeros equals the highest degree of the polynomial (but sometimes a zero will repeat). You may also see complex numbers as the zeros. So, f x = ax 2 bxc will have 2 zeros f x = ax 4 bx 3 cx 2 dx e will have 4 zeros and so on... We can find the zeros of a polynomial by factoring the original form into linear pieces like: x2 ab xab = x a x b Sometimes factoring does not work well, and we use formulaic constructions like: if f x = ax 2 bxc = 0 x= b ± b2 4ac 2a Quadratic Formula Sometimes we have to estimate zeros using numerical approximations, noting when f(x) changes in value from negative to positive or positive to negative. Let's review some of the algebraic techniques for finding roots. Theorem: If 2 quantities a and b are multiplied together and form 0, so ab = 0, then either a = 0 or b = 0. Example: 70=0 banana0=0 We use this idea when we factor to find zeros. Example: f x = x2 7x 10 Note: force leading coefficient equal to 1 by dividing Write out possible / simple arithmetic factors of the constant: 10 : {1 , 2 , 5 , 10 } Ask yourself if you can add (or subtract) any of these together to get middle number. In this case, yes 25 = 10 25=7 or 25 = 10 2 5 = 7 Form Factorization: f x = x 5 x 2 Find zeros: Let f(x) = 0 x 5 x 2 = 0 x 5 = 0 x=5 x 2 = 0 x=2 Zeros of f x = {2 , 5} = x Theorem: If p(x) is a polynomial, then: p x = x a q x for some polynomial q(x), if and only if x = a is a zero of p(x) (i.e.: p a = 0 ) Example: Sow that x = 3 is a zero of f x = 6x 3 19 x 2 x6 If x = 3, then f (3) = 0. So, test: f 3 = 6 33 19 32 3 6 Yeah! = 6 2719 93 6 = 0 Now, we can use this known zero to help us find other zeros: Right now we know that (x-3) is a linear factor of f(x). So f x = x 3 q x . We just need to figure out q(x). Option 1: Long Division: 6x 2 x 2 x 3 6x 3 19x2 x 6 6x 3 18 x 2 x2 x 6 x 2 3x 2x 6 2x 6 0 So: f x = x 3 6x 2 x2 6x 2 x2 = 0 3x 2 2x1 = 0 3x 2 = 0 2x1 = 0 2 1 x= x = 3 2 x 3 = 0 x= 0 { Zeros of f x = x = 1 2 , ,3 2 3 } Option 2: Synthetic Division p (fully reduced) is q a zero (when said polynomial has integer coefficients) then p is a factor of the constant term and q is a factor of the leading coefficient. Rational Zero Theorem: if f x = polynomial and the rational number Synthetic Review: f x = 6x 3 19x2 3 6 -19 3 6 6 -1 Known Zero x 6 Coefficient of each 3 1 3 2 term in decreasing order 1 6 -2 0 Remainder Note: We want the remainder to be zero. This technique is particularly helpful for high degree polynomials. Yielding: f x = x 3 6x 2 x2 Now we manage q x = 6x 2 x 2 From the leading coefficient 6 = {1 , 2 , 3 , 6} q constant 2 = {1 , 2 } p Possible Roots: ± 1 1 1 2 2 2 2 1 ,± ,± ,± ,± ,± ,± ,± 1 2 3 6 1 2 3 6 Test: 1/1 = 1 6 -1 -2 16 1 5 6 5 3 Note: the remainder is NOT zero, 1 is not a factor. 1/ 2 Aha! 6 -1 -2 1 6 2 1 4 2 6 -4 0 q x = x 1 6x 4 2 6x 4 = 0 2 x = 3 1 x = 2 as before Option 3: Quadratic Formula We realize: f x = x 3 6x 2 x2 We assume 6x 2 x 2 = 0 ax 2 bxc = 0 So: a = 6, b = -1, c = -2 and apply form: x= 1± 12 4 62 b ± b2 4ac 1± 148 1±17 = = = 2a 26 12 12 17 8 2 17 6 1 = = ; = = 12 12 3 12 12 2 { Zeros of f x = x = } 1 2 , , 3 as before. 2 3 NOTE: This is the most robust method for 2nd degree polynomials but other preceding methods work for any degree.