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4-5, 4-6 Factor and Remainder Theorems

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r is an x intercept of the graph of the function If r is a real number that is a zero of a function then x = r is a solution, or root, of the equation x - r is a factor of the polynomial

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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

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To find the zeros, solve the equation when Begin with quadratics. Techniques for solving quadratic equations Factor, set factors = to zero, solve Complete the square Use quadratic formula

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Reminder:

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0= x(x 3 + 27) 0=x(x 3 + 3 3 ) Use the rule a 3 + b 3 = (a + b) (a 2 – ab + b 2 ). x(x + 3)(x 2 – x 3 + 3 2 ) 0=x(x + 3)(x 2 – 3x + 9) f(x)=x 4 +27x 0=x 4 + 27x

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Roots are real and rational

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The multiplicity of root r is the number of times it occurs as a root. When a real root has odd multiplicity greater than 1, the graph “bends” as it crosses the x-axis. When a real root has even multiplicity, the graph of y = P(x) touches the x-axis but does not cross it. -2 occurs twice as a root, so it has a multiplicity of 2

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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.

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2 is a double root, real, and rational

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4.3.7

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Roots are real but irrational

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In all of the previous 3 examples, the roots were real, they were all x- intercepts, places where the graph crossed the x axis. Now consider the function

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Roots are imaginary.

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Roots are imaginary, graph does NOT cross the x axis.

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What about other polynomials, not quadratic? f(x)=4x 4 + 108x f(x)x 4 + 25 = 26x 2 f(x)= 2x 6 – 10x 5 – 12x 4 = 0 Where to start?

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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).

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Divide using synthetic division. (3x 4 – x 3 + 5x – 1) ÷ (x + 2) Step 1 Find a. Use 0 for the coefficient of x 2. For (x + 2), a = –2. a = –2 3 – 1 0 5 –1–2 Step 2 Write the coefficients and a in the synthetic division format.

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Draw a box around the remainder, 45. 3 –1 0 5 –1–2 Step 3 Bring down the first coefficient. Then multiply and add for each column. –6 345 Step 4 Write the quotient. 3x 3 – 7x 2 + 14x – 23 + 45 x + 2 Write the remainder over the divisor. 46–2814 –2314–7

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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.

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Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x 3 + 5x 2 – x + 7 for x = 2. 2 5 –1 7 2 4 241 P(2) = 41 Check Substitute 2 for x in P(x) = 2x 3 + 5x 2 – x + 7. P(2) = 2(2) 3 + 5(2) 2 – (2) + 7 P(2) = 41 3418 179

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Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 6x 4 – 25x 3 – 3x + 5 for x = –. 6 –25 0 –3 5 –2 67 1 3 – 1 3 P( ) = 7 1 3 2–39 –69–27

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The Remainder Theorem states that if a polynomial is divided by (x – a), the remainder is the value of the function at a. So, if a is a zero, then (x – a) is a factor of P(x), then P(a) = 0. 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.

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Determine whether the given binomial is a factor of the polynomial P(x). A. (x + 1); (x 2 – 3x + 1) Find P(–1) by synthetic substitution. 1 –3 1–1 15–4 4 B. (x + 2); (3x 4 + 6x 3 – 5x – 10) Find P(–2) by synthetic substitution. 3 6 0 –5 –10–2 –6 30 1000 –500

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5-Minute Check Lesson 4-4A

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Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation.

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Lesson Overview 4-4A

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5-Minute Check Lesson 4-5A

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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. noyes There are 4 changes. Therefore there are 4, 2, or 0 positive real roots.

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Descartes’ Rule of Signs 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. no yes There is 1 change. Therefore there is 1 negative real root. The rule can also be applied to find the number of negative real roots. First find f(-x) and count the sign changes. no

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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 2x 3 – 9x 2 + 2 = 0. By the Fundamental Theorem of Algebra, we know that there are 3 complex roots. The Rational Root Theorem identifies possible rational roots. ±1, ±2 = ±1, ±2, ±. 1 2 p = 2 and q = 2

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Lesson Overview 4-4A

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5-Minute Check Lesson 4-5A

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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 2x 3 – 9x 2 + 2 = 0. By the Fundamental Theorem of Algebra, we know that there are 3 complex roots. The Rational Root Theorem identifies possible rational roots. ±1, ±2 = ±1, ±2, ±. 1 2 p = 2 and q = 2

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Applying Descartes’ Rule of Signs 2x 3 – 9x 2 + 2 = 0. There are 2 or 0 positive real roots -2x 3 – 9x 2 + 2 = 0. There is 1 negative real root. Graph y = 2x 3 – 9x 2 + 2 to find the x-intercepts.

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2 –9 0 2 1 20 –2 –4 –8 Test the possible rational root. 1 2 1 2 Test. The remainder is 0, so (x – ) is a factor. 1 2 1 2 Is the Depressed Equation. remainder constant coefficient of x 2 coefficient of x

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The polynomial factors into (x – )(2x 2 – 8x – 4). 1 2 Solve 2x 2 – 8x – 4 = 0 to find the remaining roots. 2(x 2 – 4x – 2) = 0 Factor out the GCF, 2 Use the quadratic formula to identify the irrational roots. 4± 16 + 8 2 6 2 x The fully factored equation is 1 2 x – x – 2 + 6 x – 2 – 6 = 0 2 The roots are,, and. 1 2 2 6

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Identify all the real roots of 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 Applying Descartes’ Rule of Signs 3 or 1 positive real roots 1 negative real root

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Graph the function to get a starting place. Use synthetic substitution and the factor theorem to test for roots 1 -7 13 3 -18 1 -8 21 -18 0 -1 8 -21 18 21 -8 21 -18 1 -6 9 0 2 -12 18

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We are left with the depressed equation This quadratic factors easily into Therefore 3 is a double root. Finally the roots are -1, 2 and 3 (double). The factors are

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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 x 3 – x 2 – 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.

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1. 5x 4 – 20x 3 + 20x 2 = 0 4. x 3 + 9 = x 2 + 9x 4 with multiplicity 32. x 3 – 12x 2 + 48x – 64 = 0 –3, 3, 1 0 and 2 each with multiplicity 2 3. Identify all the real roots of x 3 + 5x 2 – 3x – 3 = 0. Identify the roots of each equation. State the multiplicity of each root. 1, 3 + 6, 3 6

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