A repeated root is a root that occurs more than once in a polynomial function.

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1 Unit 2A, Lesson 3.3 Finding Zeros Synthetic division, along with your knowledge of end behavior and turning points, can be used to identify the x-intercepts of a polynomial function. This information allows for more accurate sketches of functions. Recall that roots are the x-intercepts of a function. In other words, these are the x-values for which a function equals 0. These zeros are another way of referring to the roots of a function. When a polynomial equation with a degree greater than 0 is solved, it may have one or more real solutions, or it may have no real solutions (in which case it would have complex solutions). Recall that both real and imaginary numbers belong to the set of complex numbers; therefore, all polynomial functions with a degree greater than 0 will have at least one root in the set of complex numbers. This is referred to as the Fundamental Theorem of Algebra. Fundamental Theorem of Algebra If p(x) is a polynomial function of degree n 1 with complex coefficients, then the related equation p(x) = 0 has at least one complex solution (root). A repeated root is a root that occurs more than once in a polynomial function. Recall that the solutions to a quadratic equation that contains imaginary numbers come in pairs. These are called complex conjugates, the complex number that when multiplied by another complex number produces a value that is wholly real; the complex conjugate of a + bi is a bi. If an imaginary number is a zero of a function, its conjugate is also a zero of that function. This is true for all polynomial functions, and is known as the Complex Conjugate Theorem. Complex Conjugate Theorem Let p(x) be a polynomial with real coefficients. If a + bi is a root of the equation p(x) = 0, where a and b are real and b 0, then a bi is also a root of the equation. For a polynomial function p(x), the factor of a polynomial is any polynomial that divides evenly into that function. Recall that when a polynomial is divided by one of its factors, there is a remainder of 0 and the result is a depressed polynomial. This is an illustration of the Factor Theorem. Factor Theorem The binomial x a is a factor of the polynomial p(x) if and only if p(a) = 0, where a is a real number. The Factor Theorem can help to find all the factors of a polynomial. To do this, first show that the binomial in question is a factor of the polynomial. If the remainder is 0, then the binomial is a factor. Then, determine if the resulting depressed polynomial can be factored. The identified factors indicate where the function crosses the x-axis. The zeros of a function are related to the factors of the polynomial. The graph of a polynomial function shows the zeros of the function, which are the x-intercepts of the graph. It is often helpful to know which integer values of a to try when determining p(a) = 0. Unit2ALesson /24/2018

2 Use the Integral Zero Theorem to determine the zeros of a polynomial function. Identify the factors of the constant term of a polynomial function and use substitution to determine if each number results in a zero. Use synthetic division to determine the remaining factors. Integral Zero Theorem If the coefficients of a polynomial function are integers such that a n 1 and a 0 0, then any rational zeros of the function must be factors of a 0. For example, consider the equation p(x) = x x a n 1 and a A possible zero of this function is 5 because 5 is a factor of 25. The zeros of any polynomial function correspond to the x-intercepts of the graph and to the roots of the corresponding equation. If a polynomial has a factor x a that is repeated n times, then the root is called a repeated root and x = a is a zero of multiplicity. Multiplicity refers to the number of times a zero of a polynomial function occurs. If the multiplicity is odd, then the graph intersects the x-axis at the point (x, 0). If the multiplicity is even, then the graph just touches the axis at the point (x, 0). If p(x) is a polynomial with real coefficients whose terms are arranged in descending powers of the variable, then the number of positive zeros of y = p(x) is the same as the number of sign changes of the coefficients of the terms, or is less than this by an even number. Recall that when solving a quadratic function, we used the quadratic formula, which generated pairs of roots. Often, these roots were complex and the x-intercepts could not be graphed on the coordinate plane. For this reason, we must count down from our maximum number of zeros by twos to determine the number of real zeros. For example, for a polynomial with 3 sign changes, the number of positive zeros may be 3 or 1, since 3 2 = 1. Also, the number of negative zeros of y = p(x) is the same as the number of sign changes of the coefficients of the terms of p( x), or is less than this number by an even number. Common Errors/Misconceptions not finding all the factors of a polynomial function making sign errors when performing synthetic division misusing the terms roots and factors Unit2ALesson /24/2018

3 Example 1 Given the equation x 3 + 4x 2 3x 18 = 0, state the number and type of roots of the equation if one root is Use synthetic division to find the depressed polynomial. The related polynomial is x 3 + 4x 2 3x 18, with coefficients 1, 4, 3, and 18. One root of the equation is 3 ; therefore, a factor of the related polynomial is [x ( 3)], which simplifies to x + 3. Divide the related polynomial by x + 3 to find the depressed polynomial. Let the value of a be The depressed polynomial is x 2 + x Factor the depressed polynomial to find the remaining factors. Use previously learned strategies to factor the polynomial. x 2 + x 6 Depressed polynomial (x 2)(x + 3) Factor the polynomial The remaining factors are (x 2) and (x + 3). 3. State the number of roots and type of roots of the equation. Example 2 The factors of the related polynomial are (x + 3), (x 2), and (x + 3). Recall that you divided the depressed polynomial by (x + 3) before finding the remaining factors, so you must include (x + 3) as a factor. Therefore, the roots of the equation are 3, 3, and 2. Since 3 appears twice, it is a repeated root. Because of the repeated root, this equation has only 2 real roots: 3 and 2. x 3 is said to be a zero of multiplicity 2. Find all the zeros of the function f(x) = x 3 + x x Verify the zeros by creating a graph. 1. Determine the possible number of zeros. The term with the highest power is x 3, so f(x) has a degree of 3; therefore, it has 3 zeros. 2. Determine the type of zeros. Identify the signs for each term of the function f(x) = x 3 + x x Notice that there are no sign changes between each term; therefore, the function f(x) has no positive real zeros. Find f( x) to determine the number of negative real zeros. f(x) = x 3 + x x + 51 f( x) = ( x) 3 + ( x) ( x) + 51 f( x) = x 3 + x 2 11x + 51 Original Function Substitute x for x Simplify Identify the signs for each term of the function f( x), and count the number of sign changes. f( x) = x 3 + x 2 11x + 51 Unit2ALesson /24/2018

4 Example 2 (continued) There are a total of three sign changes, so the function could have 3 negative real zeros. However, recall that the number of negative zeros is either the number of sign changes of f( x), or the number of sign changes less an even number. In this case, 3 2 = 1, so the function could have 1 negative zero. The function f(x) has either 3 real zeros, or 1 real zero and 2 imaginary zeros. 3. Find the possible zeros of the function. You have already determined that none of the zeros are positive. You can determine from the equation that f(0) = 51, since f(0) = (0) 3 + (0) 2 +11(0) + 51 = 51. Evaluate f(x) for negative integral values from 4 to 1 using synthetic division. Record each value of x, the coefficients of the depressed polynomial, and the remainder in a table x Notice that one zero occurs at x = 3. From the coefficients in the table, you can determine that the depressed polynomial of this zero is x 2 2x The depressed polynomial is already written in standard form, ax 2 + bx + c. Use the quadratic formula to find the zeros of the related equation, x 2 2x + 17=0. Let a = 1, b = 2, and c = b b 4ac x Quadratic formula 2a 2 ( 2) ( 2) 4(1)(17 ) x Substitute 1 for a, 2 for b, and 17 for c 2(1) 2 x x 2 2 8i x 2 x 1 4i The function f(x) = x 3 + x x + 51 has one real zero at x 3 and two imaginary zeros at x 1 4i and x 1 4i. Unit2ALesson /24/2018

5 Example 2 (continued) 4. Graph the function f(x) = x 3 + x x Use a table of values to sketch the graph, or use a graphing calculator to create a graph of the function. To graph a function using a graphing calculator, follow these general steps for your calculator model. On a TI-83/84: Step 1: Press [ Y = ]. Step 2: Type the function into Y 1, or any available equation. Use the [ X, T, ϴ, n ] button for the variable x. Use [ ^ ] to enter powers greater than 2; use the [ x 2 ] button for a square. Step 3: Press [ WINDOW ]. Enter values for Xmin, Xmax, Ymin, and Ymax in relation to the change in the viewing window. The Xscl and Yscl are arbitrary. Leave Xres = 1. Step 4: Press [ GRAPH ]. The following is the graph of f(x) = x 3 + x x y x Notice that the graph of the function crosses the x-axis at only one point ( 3, 0). This verifies there is only one real root. Unit2ALesson /24/2018

6 Example 3 Write the simplest polynomial function with integral coefficients that has the zeros 5 and 3 i. 1. Determine additional zeros of the function. Because 3 i is a zero, then according to the Complex Conjugate Theorem, the conjugate 3 + i is also a zero. 2. Use the zeros to write the polynomial as a product of the factors. The zeros are 5, 3 + i, and 3 i. These can be written as the factors x 5, x (3 + i ), and x (3 i ). The polynomial function written in factored form with zeros 5, 3 + i, and 3 i, is f(x) = (x 5) [ x (3 + i ) ] [ x (3 i ) ]. 3. Multiply the factors to determine the polynomial function. f(x) = (x 5) [ x (3 + i ) ] [ x (3 i ) ] f(x) = (x 5) [ (x 3) i ] [ (x 3) + i ] f(x) = (x 5) [ (x 3) 2 i 2 ] f(x) = (x 5) ( x 2 6x + 9 i 2 ) Polynomial function written in factored form Regroup terms Rewrite as the difference of two squares Simplify f(x) = (x 5) [ x 2 6x + 9 ( 1 ) ] Replace i 2 with 1, since i 2 = 1 f(x) = (x 5) ( x 2 6x + 10 ) f(x) = x 3 6x x 5x x 50 f(x) = x 3 11x x 50 Simplify Distribute Combine like terms The polynomial function of least degree with integral coefficients whose zeros are 5 and 3 i is f(x) = x 3 11x x 50. Unit2ALesson /24/2018