Theorems about Roots of Polynomial Equations and The fundamental theorem of Algebra What you’ll learn To solve equations using the Rational Root Theorem. To use the conjugate Root Theorem. To use the Fundamental Theorem of Algebra to solve polynomial equations with complex solutions Vocabulary Rational Root Theorem. Conjugate Root Theorem. Descartes’ Rule of Signs. Fundamental theorem of Algebra.

Rational Root Theorem Example: Factoring a polynomial can be challenging, but there is a theorem to help you with that. Rational Root Theorem Example: Factors of the constant term: Factors of the leading coefficient:

Problem 1: Finding a Rational Root What are the rational roots of Leading coefficient 2 (±1,±2) and constant term 5 (±1, ±5) The only possible rational roots have the form The only possible roots are x 1 -1 5 -5 1/2 -1/2 5/2 -5/2 P(x) Note: the Rational Root Theorem does not necessarily give the zeros of the equation. It provides a list of first guesses to test as roots

What are the rational roots of Your turn What are the rational roots of Leading coefficient 1 (±1) and constant term 10 (±1, ±2,±5±10) The only possible rational roots have the form x 1 -1 2 -2 5 -5 10 -10 P(x)

Problem 2: Using the Rational Root Theorem Once you find one root, use synthetic division to factor the polynomial. Continue finding roots and dividing until you have a second degree polynomial. Use the Quadratic Formula to find the remaining roots. Problem 2: Using the Rational Root Theorem Leading coefficient 15 (±1,±3,±5,±15) and constant term 2(±1, ±2) The only possible rational roots have the form Test each possible rational root in until you find a root Test: x=1

So 2 is a root, now factor the polynomial using synthetic division

Your turn: Answer: 2,-1,-3/2 Remember steps to find rational roots. Get the constant term factors and the leading coefficient of the polynomial. 2. Find all possible rational roots 3.Test each possible rational root until you find a root. 4. Factor the polynomial until you get a quadratic. (you can use synthetic division, quadratic formula, a calculator, or any other method).when using TI-84 or 85 store the polynomial in using the Y= menu. Store the root to be tested in x(enter the number then press the key STO). Use VARS Y-VARS 1: Function to evaluate. Your turn: Answer: 2,-1,-3/2

Conjugate Root Theorem: If P(x) is a polynomial with Did you remember what is a conjugate number? If a complex number or an irrational number is a root of a polynomial equation with rational coefficients, so is its conjugate. Conjugate Root Theorem: If P(x) is a polynomial with rational coefficients, then the irrational roots of P(x)=0 that have a form occur in conjugate pairs. That is, If is an irrational root with rational then is also a root. If P(x) is a polynomial with real coefficients, then the complex roots of P(x)=0 occur in conjugate pairs. That is is a complex root with real, then

Answer Answers: Your turn Problem 3:Using the Conjugate Root Theorem to identify Roots. Answers: Remember all rational numbers are real numbers. Your turn Answer

Problem 4: Using the Conjugates to construct A Polynomial Answer: Answer:

Descartes’ Rule of Signs Theorem: Let P(x) be a polynomial with real coefficients written in standard form. *The number of positive real roots of P(x) is either equal to the number of sign changes between consecutive coefficients of P(x) or is less than that by an even number. *The number of negative real roots of P(x)=0 is either equal to the number of sign changes between consecutive coefficients of P(-x) or is less than that by an even number. In both cases, count multiple roots according to their multiplicity This theorem implies that two tests must be done on a polynomial function to determine both positive and negative real roots.

Problem 5: Using the Descartes’ Rule of Signs. There are two sign changes, + to – and - to +. Therefore, there are either 0 or 2 positive real roots. To find the negative real root you need to plug in -x into the polynomial There is only one negative real root Graph the function and recall that cubic functions have zero or two turning points. Because the graph already shows two turning points, it will not change the direction again. So there are no positive real roots.

Your turn Take a note: the degree of a polynomial equation tells Answers: there are 3 or 1 positive real roots and 1 negative root. The graph confirm one negative and one positive real Root Real roots can be confirmed graphically because they are x-intercepts. Complex roots cannot be confirmed graphically because they have an imaginary component. Take a note: the degree of a polynomial equation tells you how many roots the equation has. That is the result of the Fundamental theorem of Algebra provided by the German mathematician Carl Friedrich Gauss (1777-1855)

The fundamental Theorem of Algebra: If P(x) is a polynomial of degree n≥1, then P(x)=0 has exactly n roots, including multiple and complex roots Problem 6: Using the Fundamental Theorem of Algebra I need to find the zeros(5) and using the rational root and factor theorem, synthetic division and factoring

The factors are -4 and 1 Answers: Your turn Answers: 0,1,-5,2 Since there is no constant term, make the equation equal to zero and factor x from the polynomial. CF, GC, SD and Factoring. Answers: 0,1,-5,2

Problem 7: Finding all the zeros of a Polynomial Function Step 1: use the graphing calculator to find any real roots Step 2: Factor out the linear factors (x-3)(x+3).Use Synthetic division twice. Step3: use the quadratic formula to find the complex roots Step 4:

Your turn Answer: The Fundamental Theorem of Algebra: Here are equivalent ways to state the fundamental theorem of Algebra. You can use any of these statements to prove the others. *Every polynomial equation of degree n≥1 has exactly n roots, including multiple and complex roots. *Every polynomial of degree n≥1 has n factors. *Every polynomial function of degree n≥1 has at least one complex zero.

Classwork odd Homework even TB pg 316 exercises 9-41 and pg322 exercises 8-37