Solving Polynomial Functions Factoring a Polynomial • Remember Three Types of Problems – Difference/Sum of Two Cubes – Grouping – Quadratic Like Practice Problems • 1. 2. 3. Factor and solve x3 – 64 x3 + 6x2 -4x -24=0 4w4 + 40w2- 44=0 Using Division to Find a Zero • If you know one zero, you can use division to find another. • Remainder Theorem- If a polynomial f(x) is divided by (x-k) then the remainder is r=f(x) • Two forms of polynomial division: long division and synthetic division Long Division- can be used on any polynomial • Divide f(x)=3x4 – 5x3 + 4x – 6 by (x2 -3x + 5) x2 -3x + 5 3x4 – 5x3 + 0x2 + 4x – 6 Practice • Divide f(x)=x3 + 5x2 – 7x + 2 by x-2 Synthetic Division • Can be used to divide any polynomial by a divisor of the form x – k • To set up synthetic division, list the coefficients in a row. F(x)= 2x3 + x2 -8x + 5 by x + 3 -3 2 1 -8 5 Factor Theorem- a polynomial f(x) has a factor x-k if f(k)=0 Factor the polynomial F(x) = 3x3 – 4x2 – 28x -16 completely given that x+ 2 if a factor. The profit P ( in millions of dollars) for a shoe manufacturer can be modeled by P= -21x3 + 46x where x is the number of shoes produced (in millions). The company produces 1 million shoes and makes a profit of 25,000,000 but would like to cut back on production. What lesser number of shoes could the company make and still make the same profit? Rational Zero Theorem • One way to narrow down the possible zeros of a function is to use the Rational Zero Theorem. • If f(x)= anxn + …anx + a0 has integer coefficients, then every rational zero of f has the following form p = factors of constant term a0 q factors of leading coefficient an List the possible rational zeros of f using rational zero theorem. F(x) = x3 + 2x2 – 11x - 12 Find all real zeros of f(x)= x3 – 8x2+ 11x + 20 Find all real zeros of 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 10x4 - 11x3 – 42x2+ 7x + 12 10 -11 -42 7 12 Fundamental Theorem of Algebra 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. Complex Conjugate Theorem • If f is a polynomial function with real coefficients and a + bi is an imaginary zero of f, then a- bi is a zero of f. 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. Find the zeros F(x) = x5 – 4x4 + 4x3 + 10x2 – 13x - 14 Write a polynomial function f of least degree that has rational coefficients , a leading coefficient of 1 and 3 and 2 + √5 as zeros. Set up factors F(x)= (x – 3)(x – (2 + √5 ))(x – (2 - √5 ) Write the polynomial function f of least degree that has rational coefficients a leading coefficient of 1 and the given zeros 1. -1, 2, 4 2. 3, 3-i Descartes Rule of Signs 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. Determine the possible numbers of positive real zeros, negative real and imaginary zeros for the function F(x)= x6 – 2x5 + 3x4 – 10x3 – 6x2 -8x -8 Pos. real zeros Neg, real zeros Imag. zeros Total zeros

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# Finding Zeros of Polynomial Review

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