1. The Rational Zero Theorem The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Zero Theorem If f (x)  a n x n  a n-1 x n-1  …  a 1 x  a 0 has integer coefficients and (where is reduced) is a rational zero, then p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n. The Rational Zero Theorem If f (x)  a n x n  a n-1 x n-1  …  a 1 x  a 0 has integer coefficients and (where is reduced) is a rational zero, then p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n.

2. EXAMPLE: Using the Rational Zero Theorem Solution The constant term is –2 and the leading coefficient is 15. Divide  1 and  2 by  1. Divide  1 and  2 by  3. Divide  1 and  2 by  5. Divide  1 and  2 by  15. There are 16 possible rational zeros. The actual solution set to f (x)  15x 3  14x 2  3x – 2 = 0 is {-1,  1 / 3, 2 / 5 }, which contains 3 of the 16 possible solutions. List all possible rational zeros of f (x)  15x 3  14x 2  3x – 2.

3. EXAMPLE: Solving a Polynomial Equation Solve: x 4  6x 2  8x + 24  0. Solution Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.

4. EXAMPLE: Solving a Polynomial Equation Solve: x 4  6x 2  8x + 24  0. Solution The graph of f (x)  x 4  6x 2  8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. The zero remainder indicates that 2 is a root of x 4  6x 2  8x + 24  0. 2 10  6  8 24 2 4  4  24 1 2  2  120 2 2 8 12 1 4 6 0 The zero remainder indicates that 2 is a root of x 3  2x 2  2x  12 = 0.

5. EXAMPLE: Solving a Polynomial Equation Solve: x 4  6x 2  8x + 24  0. Solution Now we can solve the original equation as follows. x 4  6x 2  8x + 24  0 This is the given equation. (x – 2)(x – 2)(x 2  4x  6)  0 This was obtained from the first,second synthetic division. x – 2  0 or x – 2  0 or x 2  4x  6  Set each factor equal to zero. x  2 x  2 x 2  4x  6  Solve.

6. EXAMPLE: Solving a Polynomial Equation Solve: x 4  6x 2  8x + 24  0. Solution We can use the quadratic formula to solve x 2  4x  6  Let a  1, b  4, and c  6. We use the quadratic formula because x 2  4x  6  cannot be factored. Simplify. Multiply and subtract under the radical. The solution set of the original equation is { 2, 2,  2  i  2  i }.

7. EXAMPLE: Using the Rational Zero Theorem Solution The constant term is –1 and the leading coefficient is 2. Divide  1 by  1 Divide  1 by  2 There are 4 possible rational zeros. The actual solution set to 2x 4  3x 3  2x 2 – 1 = 0 is {-1,1  1 / 2, 1 / 2 }, which contains 2 of the 4 possible solutions. possible rational zeros

8. EXAMPLE: Solving a Polynomial Equation Solution The graph of f (x)  is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. The zero remainder indicates that -1 is a root of 2x 4  x 3  2x 2 -1  0. 2 3 2 0  -2 -1 -1 1 21 1 -1 0 1 1 1 2 2 2 0 The zero remainder indicates that1/2 is a root of 2x 3  x 2  x -1  0

9. EXAMPLE: Solving a Polynomial Equation Solve: Solution Now we can solve the original equation as follows. This is the given equation. (x +1)(x – )(2x 2  2x  2)  0 This was obtained from the first,second synthetic division. x +1  0 or x –  0 or 2x 2  2x  2  Set each factor equal to zero. x  -1 x  x 2  x  1  Solve.

10. EXAMPLE: Solving a Polynomial Equation Solution We can use the quadratic formula to solve x 2  x  1  Let a  1, b  1, and c  1. We use the quadratic formula because x 2  x  1  cannot be factored. Multiply and subtract under the radical. The solution set of the original equation is { -1,, }

11. 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a  b i is a root of a polynomial equation (b  0), then the non-real complex number a  b i is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a  b i is a root of a polynomial equation (b  0), then the non-real complex number a  b i is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. Properties of Polynomial Equations

12. If f (x)  a n x n  a n  1 x n  1  …  a 2 x 2  a 1 x  a 0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f (  x) or is less than that number by an even integer. If f (  x) has only one variation in sign, then f has exactly one negative real zero. If f (x)  a n x n  a n  1 x n  1  …  a 2 x 2  a 1 x  a 0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f (  x) or is less than that number by an even integer. If f (  x) has only one variation in sign, then f has exactly one negative real zero. Descartes' Rule of Signs

13. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x)  x 3  2x 2  5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (  x). We obtain this equation by replacing x with  x in the given function. f (  x)  (  x) 3  2(  x) 2  x  4 f (x)  x 3  2x 2  5x + 4 This is the given polynomial function. Replace x with  x.   x 3  2x 2  5x + 4

14. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x)  x 3  2x 2  5x + 4. Solution Now count the sign changes. There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3  2  1 negative real zero. f (  x)   x 3  2x 2  5x + 4 1 2 3

15. 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a  b i is a root of a polynomial equation (b  0), then the non-real complex number a  b i is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a  b i is a root of a polynomial equation (b  0), then the non-real complex number a  b i is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. Properties of Polynomial Equations

16. Complex Conjugates Theorem Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs. If a + bi (b ≠ 0) is a zero of a polynomial function, then its Conjugate, a - bi, is also a zero of the function.

17. Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Theorem. Ex: Find all the roots of If one root is 4 - i. Because of the Complex Conjugate Theorem, we know that another root must be 4 + i. Can the third root also be imaginary?

18. Example (con’t) Ex: Find all the roots of If one root is 4 - i. If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)]. Multiply these factors:

19. Example (con’t) Ex: Find all the roots of If one root is 4 - i. If the product of the two non-real factors is then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by : The third root is x = -3 The answer of x 3 -5x 2 -7x+ 15 =0 is {4-i,4+i,-3}

20. So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i. Let’s find such a polynomial by putting the roots in factor form and multiplying them together. Multiply the last two factors together. All i terms should disappear when simplified. Now multiply the x – 2 through Here is a 3 rd degree polynomial with roots 2, 1 - 3i and 1 + 3i If x = the root then x - the root is the factor form.

21. Conjugate Pairs Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a – bi is also a zero of the function (the polynomial function must have real coefficients) EXAMPLES: Find the polynomial equation with the given zeros -1, -1, 3i, -3i 2, 4 + i, 4 – i

22. 0= (x-1)(x-(-2+i))(x-(-2-i)) 0= (x-1)(x+2 - i)(x+2+ i) 0= (x-1)[(x+2) - i] [(x+2)+i] 0= (x-1)[(x+2) 2 - i 2 ] Foil 0=(x-1)(x 2 + 4x + 4 – (-1))Take care of i 2 0= (x-1)(x 2 + 4x + 4 + 1) 0= (x-1)(x 2 + 4x + 5)Multiply 0= x 3 + 4x 2 + 5x – x 2 – 4x – 5 0= x 3 + 3x 2 + x - 5 Now write a polynomial equation of least degree that has real coefficients, a leading coeff. of 1 and 1, -2+i, -2-i as zeros.

23. Note: 2+i means 2 – i is also a zero 0= (x-4)(x-4)(x-(2+i))(x-(2-i)) 0= (x-4)(x-4)(x-2-i)(x-2+i) 0= (x 2 – 8x +16)[(x-2) – i][(x-2)+i] 0= (x 2 – 8x +16)[(x-2) 2 – i 2 ] 0= (x 2 – 8x +16)(x 2 – 4x + 4 – (– 1)) 0= (x 2 – 8x +16)(x 2 – 4x + 5) 0= x 4 – 4x 3 +5x 2 – 8x 3 +32x 2 – 40x+16x 2 – 64x+80 0= x 4 -12x 3 +53x 2 -104x+80 Now write a polynomial equation of least degree that has real coefficients, a leading coeff. of 1 and 4, 4, 2+i as zeros.