 # Pre-Calculus For our Polynomial Function: The Factors are:(x + 5) & (x - 3) The Roots/Solutions are:x = -5 and 3 The Zeros are at:(-5, 0) and (3, 0)

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Pre-Calculus

For our Polynomial Function: The Factors are:(x + 5) & (x - 3) The Roots/Solutions are:x = -5 and 3 The Zeros are at:(-5, 0) and (3, 0)

The only way that x 2 +2x - 15 can = 0 is if x = -5 or x = 3 4 Rearrange the terms to have zero on one side: 4 Factor: 4 Set each factor equal to zero and solve:

The points where y = 0 are called the x-intercepts of the graph. The x-intercepts for our graph are the points... and (-5, 0) (3, 0)

If a polynomial has ‘n’ complex roots will its graph have ‘n’ x-intercepts? In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are 3 x-intercepts.

We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring. Some are easier to factor than others! The roots are: 0, -2, 2

Just because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real! In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components.

Find the zeros of 2, -3 (d.r), 1, -4 EXAMPLE 1

Solveby factoring. Answer: The solution set is {0, –4}. Original equation Add 4x to each side. Factor the binomial. Solve the second equation. Zero Product Property or

Solveby factoring. Original equation Subtract 5x and 2 from each side. Factor the trinomial. Zero Product Property or Solve each equation. Answer: The solution set is Check each solution.

Answer: The solution set is {3}. Solveby factoring. Original equation Add 9 to each side. Factor. Zero Product Property or Solve each equation.

Check The graph of the related function, intersects the x -axis only once. Since the zero of the function is 3, the solution of the related equation is 3.

Factor Theorem 1. If f(c)=0, that is c is a zero of f, then x - c is a factor of f(x). 2. Conversely if x - c is a factor of f(x), then f(c)=0.

Example – Find the roots/zeros for this polynomial Given -2 is a zero, use synthetic division to find the remaining roots. -2 10 9 -19 6 Don’t forget your remainder should be zero -20 22 -6 10 -11 3 0 The new, smaller polynomial is: This quadratic can be factored into: (5x – 3)(2x – 1) Therefore, the zeros to the problem

Example: Find all the zeros of each polynomial function If we were to graph this equation to check you could see the zero From looking at the graph you can see that there is a zero at -2 ZERO

We have already known that the possible rational zeros are: Find All the Rational Zeros EXAMPLE 3 So 1, 2, -2, and - 3 are rational roots

Example: The zeros of a third-degree polynomial are 2, 2, and -5. Write a polynomial. (x – 2)(x – 2)(x+5) First, write the zeros in factored form Second, multiply the factors out to find your polynomial ANSWER

Given 1 ± 3i and 2 are roots, find the remaining roots. 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. EXAMPLE 5

 f(x)= (x-1)(x-(-2+i))(x-(-2-i))  f(x)= (x-1)(x+2 - i)(x+2+ i)  f(x)= (x-1)[(x+2) - i] [(x+2)+i]  f(x)= (x-1)[(x+2) 2 - i 2 ] Foil  f(x)=(x-1)(x 2 + 4x + 4 – (-1))Take care of i 2  f(x)= (x-1)(x 2 + 4x + 4 + 1)  f(x)= (x-1)(x 2 + 4x + 5)Multiply  f(x)= x 3 + 4x 2 + 5x – x 2 – 4x – 5  f(x)= x 3 + 3x 2 + x - 5

 Note: 2+i means 2 – i is also a zero  F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i))  F(x)= (x-4)(x-4)(x-2-i)(x-2+i)  F(x)= (x 2 – 8x +16)[(x-2) – i][(x-2)+i]  F(x)= (x 2 – 8x +16)[(x-2) 2 – i 2 ]  F(x)= (x 2 – 8x +16)(x 2 – 4x + 4 – (– 1))  F(x)= (x 2 – 8x +16)(x 2 – 4x + 5)  F(x)= x 4 – 4x 3 +5x 2 – 8x 3 +32x 2 – 40x+16x 2 – 64x+80  F(x)= x 4 -12x 3 +53x 2 -104x+80

 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 a polynomial with the given zeros  -1, -1, 3i, -3i  2, 4 + i, 4 – i

If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula. We can solve the resulting polynomial to get the other 2 roots: (x - 5) is a factor

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 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. The solution set of the original equation is { 2,  2 –  i,  2  i }. Ise Quadratic Formula

(Given that 1 + 3i is a zero of f) (Given that 5 + 2i is a zero of f)

FIND ALL RATIONAL ROOTS: List all possible rational zeros of f (x)  15x 3  14x 2  3x – 2. 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.

EXAMPLE: Solving a Polynomial Equation Solve: x 4  6x 2  8x + 24  0. Solution Recall that we refer to the zeros of a polynomial function and the roots of a polynomial equation. 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. FIND ALL POSSIBLE RATIONAL ROOTS

Arrange the terms of the polynomial P(x) in descending degree: The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real Roots (or less by any even number) The number of times the coefficients of the terms of P(-x) change sign = the number of Negative Real Roots (or less by any even number) In the examples that follow, use Descartes’ Rule of Signs to predict the number of + and - Real Roots!

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

3.4: Zeros of Polynomial Functions 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. f (  x)   x 3  2x 2  5x + 4 1 2 3 There are 0 positive roots, and 3 negative roots

For higher degree polynomials, finding the complex roots (real and imaginary) is easier if we know one of the roots. Descartes’ Rule of Signs can help get you started. Complete the table below:

For the polynomial: All possible values of: All possible Rational Roots of the form p/q:

For the polynomial: Descartes’ Rule: All possible Rational Roots of the form p/q:

For the polynomial: Substitute each of our possible rational roots into f(x). If a value, a, is a root, then f(a) = 0. (Roots are solutions to an equation set equal to zero!)

Descartes’s Rule of Signs EXAMPLES: describe the possible real zeros

Find the zeros of Hint: 2 is a zero X 2 21 -13 6 2 4 5 10 -3 -6 0

Find the zeros of Hint: 4 is a zero X 4 10 -11 -20 1 4 4 16 5 20 0

Find the zeros of Hint: 2 is a zero X 2 1-2 -3 6 1 2 0 0 -6 0

Find the zeros of Hint: 2 is a zero X 2 15 -2 -24 1 2 7 14 12 24 0

No Calculator Given –2 is a zero of find ALL the zeros of the function. -2 1 -5 6 1 -2 -4 8 3 -6 0

No Calculator Given 5 is a zero of find ALL the zeros of the function. 5 1-5 -4 20 1 5 0 0 -4 -20 0 No constant, so 0 is a zero:

No Calculator Given -1 and 3 are zeros of find ALL the zeros of the function. 3 1 3 -7 -21 12 36 0 1-9 23 -3 -36 1 -10 10 33 -33 -36 36 0

No Calculator Given is a zero of find ALL the zeros of the function. 2 3 -6 -9 4 6 0 /2 2-9 13 -6 3 1 -3 2

No Calculator Given is a zero of find ALL the zeros of the function. 3 2 -6 -4 -9 -6 0 /3 3-8 -5 6 2 1 -2 -3

No Calculator Given 2 is a zero of find ALL the zeros of the function. 2 1-6 13 -10 1 2 -45 10 0 -8

No Calculator Given –3 is a zero of find ALL the zeros of the function. -3 1 3 1 3 1 0 0 1 0

No Calculator Find a polynomial function with real coefficients which has zeros of 1, -2, and 3.

No Calculator Find a polynomial function with real coefficients which has zeros of 0, 2, -2, and 5.

No Calculator Find a polynomial function with real coefficients which has zeros of 3/2, 2, and 1.

No Calculator Find a polynomial function with real coefficients which has zeros of 2 and i. If i is a root, then –i is a root as well

No Calculator Find a polynomial function with real coefficients which has zeros of 1 and 2 + i. If 2 + i is a root, then 2 – i is a root as well

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