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4-5, 4-6 Factor and Remainder Theorems If r is a real number that is a zero of a function then f x 0 •r is an x intercept of the graph of the function •x = r is a solution, or root, of the equation •x - r is a factor of the polynomial First: How many zeros are we expecting? How many were there in each of our quadratic examples? Fundamental Theorem of Algebra Every polynomial of degree n≥1 has at least one complex zero. Every polynomial of degree n≥1 has exactly n complex zeros, counting multiple roots. Complex zeros are real, imaginary or a combination To find the zeros, solve the equation when f x 0 Begin with quadratics. Techniques for solving quadratic equations •Factor, set factors = to zero, solve • Complete the square • Use quadratic formula Reminder: f(x)=x4 +27x 0=x4 + 27x 0= x(x3 + 27) 0=x(x3 + 33) x(x + 3)(x2 – x 3 + 32) Use the rule a3 + b3 = (a + b) (a2 – ab + b2). 0=x(x + 3)(x2 – 3x + 9) x0 x3 0 x 3 0 x 3x 9 2 f x x 2 x 8 2 0 x 2x 8 2 x4 x 2 f x x 2 x 8 2 15 10 fx = x2-2x-8 5 -20 -10 10 -5 -10 Roots are real and rational 20 f x x 4 x 4 2 0 x 4x 4 2 x2 x2 The multiplicity of root r is the number of times it occurs as a root. -2 occurs twice as a root, so it has a multiplicity of 2 When a real root has even multiplicity, the graph of y = P(x) touches the x-axis but does not cross it. When a real root has odd multiplicity greater than 1, the graph “bends” as it crosses the x-axis. You cannot always determine the multiplicity of a root from a graph. It is easiest to determine multiplicity when the polynomial is in factored form. f x x 4 x 4 2 10 5 x -10 10 -5 2 is a double root, real, and rational 20 f ( x) x 5 x 3 0 x 5x 3 2 2 b b 4ac x 2a 2 5 13 x 2 5 13 4.3 x 2 5 13 x 2 .7 f ( x) x 5 x 3 2 4 2 5 -2 Roots are real but irrational 10 In all of the previous 3 examples, the roots were real, they were all xintercepts, places where the graph crossed the x axis. Now consider the function f x x 2 x 5 2 f x x 2 x 5 2 b b 4ac x 2a 2 16 2 i 16 2 2 21 2i 1 2i 2 2 Roots are imaginary. 2 4i 2 f x x 2 x 5 2 12 10 8 6 4 2 -10 -5 5 10 -2 Roots are imaginary, graph does NOT cross the x axis. What about other polynomials, not quadratic? f(x)=4x4 + 108x f(x)x4 + 25 = 26x2 f(x)= 2x6 – 10x5 – 12x4 = 0 Where to start? Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a). Divide using synthetic division. (3x4 – x3 + 5x – 1) ÷ (x + 2) Step 1 Find a. a = –2 For (x + 2), a = –2. Step 2 Write the coefficients and a in the synthetic division format. –2 3 – 1 0 5 –1 Use 0 for the coefficient of x2. Step 3 Bring down the first coefficient. Then multiply and add for each column. –2 3 –1 0 5 –1 –6 14 –28 46 3 –7 14 –23 45 Draw a box around the remainder, 45. Step 4 Write the quotient. 3x3 – 7x2 45 + 14x – 23 + x+2 Write the remainder over the divisor. You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem. Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x3 + 5x2 – x + 7 for x = 2. 2 2 5 –1 2 4 9 7 18 34 17 41 P(2) = 41 Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7. P(2) = 2(2)3 + 5(2)2 – (2) + 7 P(2) = 41 Use synthetic substitution to evaluate the polynomial for the given value. 1 P(x) = 6x4 – 25x3 – 3x + 5 for x = – . 3 1 6 –25 0 –3 5 – 3 –2 9 –3 2 6 –27 P( 1 ) = 7 3 9 –6 7 The Remainder Theorem states that if a polynomial is divided by (x – a), the remainder is the value of the function at a. Remember that the value of a function is the y value at any given x value. If a is a root (or zero) of a function, then the value of the function at a is zero. So, if a is a zero, then (x – a) is a factor of P(x), then P(a) = 0. Determine whether the given binomial is a factor of the polynomial P(x). A. (x + 1); (x2 – 3x + 1) Find P(–1) by synthetic substitution. –1 1 –3 1 1 –1 –4 4 5 B. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–2) by synthetic substitution. –2 3 6 0 –5 –10 –6 0 0 3 0 0 –5 10 0 5-Minute Check Lesson 4-4A Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation. Lesson Overview 4-4A 5-Minute Check Lesson 4-5A Descartes’ Rule of Signs Suppose P(x) is a polynomial whose terms are arranged in descending powers of the variable. Then the number of positive real zeros is the same as the number of changes in sign of the coefficients of the terms or is less than this by an even number. f ( x ) 2 x 5 3x 4 6 x 3 6 x 2 8 x 3 no yes yes yes yes There are 4 changes. Therefore there are 4, 2, or 0 positive real roots. Descartes’ Rule of Signs The rule can also be applied to find the number of negative real roots. First find f(-x) and count the sign changes. f ( x ) 2 x 3x 6 x 6 x 8x 3 5 4 yes no 3 no 2 no no The number of negative real zeros is the same as the number of changes in sign of the coefficients of the terms or is less than this by an even number. There is 1 change. Therefore there is 1 negative real root. You can use a combination of the Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes’ Rule, Graphing, and Synthetic Substitution to determine all roots for any given polynomial. Identify all the real roots of 2x3 – 9x2 + 2 = 0. By the Fundamental Theorem of Algebra, we know that there are 3 complex roots. The Rational Root Theorem identifies possible rational roots. p = 2 and q = 2 ±1, ±2 = ±1, ±2, ± 1 . ±1, ±2 2 Lesson Overview 4-4A 5-Minute Check Lesson 4-5A You can use a combination of the Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes’ Rule, Graphing, and Synthetic Substitution to determine all roots for any given polynomial. Identify all the real roots of 2x3 – 9x2 + 2 = 0. By the Fundamental Theorem of Algebra, we know that there are 3 complex roots. The Rational Root Theorem identifies possible rational roots. p = 2 and q = 2 ±1, ±2 = ±1, ±2, ± 1 . ±1, ±2 2 Applying Descartes’ Rule of Signs 2x3 – 9x2 + 2 = 0. -2x3 – 9x2 + 2 = 0. There are 2 or 0 positive real roots There is 1 negative real root. Graph y = 2x3 – 9x2 + 2 to find the x-intercepts. gx = 2x3-9x2+2 3 2 1 -4 -2 2 -1 -2 -3 4 6 Test the possible rational root 1 . 2 1 2 2 –9 0 2 1 –4 –2 2 –8 –4 0 Test 1 . The remainder is 2 1 0, so (x – ) is a factor. 2 remainder constant coefficient of x coefficient of x2 2 x 8x 4 0 2 Is the Depressed Equation. 1 The polynomial factors into (x – )(2x2 – 8x – 4). 2 Solve 2x2 – 8x – 4 = 0 to find the remaining roots. 2(x2 – 4x – 2) = 0 x 4± 16+8 2 2 6 Factor out the GCF, 2 Use the quadratic formula to identify the irrational roots. The fully factored equation is 1 2 x – 2 x – 2 – x – 2 + 6 6 = 0 1 The roots are , 2 6 , and 2 6 . 2 Identify all the real roots of f ( x ) x 7 x 13x 3x 18 4 3 2 By the Fundamental Theorem of Algebra, we know that there are 4 roots. They may be real, imaginary or complex. The Rational Root Theorem identifies possible rational roots. p = 18 and q = 1 P : 1, 2, 3, 6, 9, 18 Q : 1 Applying Descartes’ Rule of Signs 3 or 1 positive real roots 1 negative real root Graph the function to get a starting place. Use synthetic substitution and the factor theorem to test for roots -1 1 1 2 1 1 -7 13 3 -18 -1 8 -21 18 -8 21 -18 0 -8 21 -18 2 -12 18 -6 9 0 We are left with the depressed equation x 6x 9 0 2 This quadratic factors easily into x 3 x 3 0 Therefore 3 is a double root. Finally the roots are -1, 2 and 3 (double). The factors are x 3 x 3 x 1 x 2 The Conjugate Pairs Theorem says that irrational roots and imaginary roots come in conjugate pairs. For example, if you know that 1 + is a root of x3 – x2 – 3x – 1 = 0, then you know that 1– is also a root. If you know that -3i is a root, then 3i is also a root. If you know that 4+2i is a root, then 4-2i is also a root. Identify the roots of each equation. State the multiplicity of each root. 0 and 2 each with 1. 5x4 – 20x3 + 20x2 = 0 multiplicity 2 2. x3 – 12x2 + 48x – 64 = 0 4 with multiplicity 3 3. Identify all the real roots of x3 + 5x2 – 3x – 3 = 0. 1, 3 + 6, 3 6 4. x3 + 9 = x2 + 9x –3, 3, 1