Solving Polynomial Functions. Factoring a Polynomial. Remember Three Types of Problems Difference/Sum of Two Cubes Grouping Quadratic Like. Practice Problems. Factor and solve x 3 – 64 x 3 + 6x 2 -4x -24=0 4w 4 + 40w 2 - 44=0. Using Division to Find a Zero.
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by (x2 -3x + 5)
x2 -3x + 5 3x4 – 5x3 + 0x2 + 4x – 6
-3 2 1 -8 5
Factor the polynomial
F(x) = 3x3 – 4x2 – 28x -16 completely given that x+ 2 if a factor.
p = factors of constant term a0
q factors of leading coefficient an
F(x) = x3 + 2x2 – 11x - 12
f(x)= 10x4 - 11x3 – 42x2+ 7x + 12
Step 1: List possible rational zeros.
Step 2: Use graphing calculator to narrow down choice\
Step 3: Use synthetic division to test zero
10 -11 -42 7 12
Theorem: If f(x) is a polynomial of degree n where n >0 then the equation f(x)=0 has at least one solution in the set of complex numbers.
Corollary: if f(x) is a polynomial of degree n where n>0 then the equation f(x)=0 has exactly n solutions provided each solution repeated twice is counted as 2 solutions, each solution repeated three times is counted as 3 solutions and so on.
Irrational Conjugate Theorem
Suppose f is a polynomial function with rational coefficients and a and b are rational numbers such that √ b is irrational. If a + √b is a zero of f , then a – √b is also a zero of f.
F(x) = x5 – 4x4 + 4x3 + 10x2 – 13x - 14
Set up factors
F(x)= (x – 3)(x – (2 + √5 ))(x – (2 - √5 )
2. 3, 3-i
Let f(x) = anxn + …anx + a0 be a polynomial function with real coefficients.
-The number of positive real zeros of f is equal to the number of changes in sign of coefficients of f(x) or is less than this by an even number.
- The number of negative real zeros of f is equal to the number of changes in sign of coefficients of f(-x) or is less than this by an even number.
F(x)= x6 – 2x5 + 3x4 – 10x3 – 6x2 -8x -8