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Published byCaleb Carter Modified over 10 years ago

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**Lesson 3.4 – Zeros of Polynomial Functions Rational Zero Theorem**

Represent a polynomial equation of degree n . If a rational number , where p and q have no common factors, is a root of the equation, then p is a factor of the constant term and q is a factor of the leading coefficient. Ex. 1 List all possible roots of Then determine the rational roots. List possible values of p: List possible values of q: Possible rational roots:

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**You Try: List all possible rational zeros of f(x) = x3 + 2x2 – 5x – 6**

Possible values of p: Possible values of q: Possible rational roots(p/q):

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**Finding Zeros of a Polynomial Function**

Now, use synthetic division to test and find the roots/factors. The last number must be a zero to show the root is a factor. Degree is 3, so there should be 3 solutions. Possible rational roots:

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**Checking with Synthetic Division**

is not a zero because the remainder does not equal 0!! Now let’s try -2. is a zero!!!

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**Finding the Zero (cont.)**

Take -2 and write it as a factor which is x+ 2 and take your answer from synthetic division and put it into a polynomial 6x2 – x -1. Now factor 6x2 – x -1 (2x – 1 )(3x + 1) Now put all the factors together (x+2)(2x-1)(3x+1). Put factors equal to zero to find the zeros. X= -2, ½, -1/3 (3 real rational solutions)

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**The process: Don’t forget:**

Step 1:Find your p’s and q’s and list all possible roots. Step 2:Number of roots/zeros is based on highest degree. Use synthetic division to find your first root. If that does not work, USE YOUR CALCULATOR!!! Remember your multiplicity ideas as well. If the polynomial crosses the x axis, the multiplicity is odd. If the polynomial touches and turns around, it is even. Step3: After finding a root, factor the rest on your own. If not factorable, use the quadratic formula. Step 4: Then, solve for the rest of the roots. Roots can be real or imaginary. If the roots are imaginary, then they occur in conjugate pairs! To set up factors (in parenthesis) just change their signs.

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**You Try!! Find all zeros of f(x) = x3 + 7x2 + 11x – 3**

Step 1 – Find possible rational roots. p: q: possible rational roots: Use synthetic division to find one rational root or by the calculator. By using the calculator, find one zero. Show on the calculator to class. Hint: You will need to use the quadratic formula One root is 3 from calculator. Now find the other roots. How many should there be? 3

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**Answer: The solution set is {-3,-2 - √5, -2+√5}**

Your solutions can be imaginary or real. If your solution is imaginary, it will be written as a complex conjugate. If it is real, it could be rational (nice numbers) or irrational (not nice numbers).

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**You Try Again: Solve: x4 + 6x3 + 22x2 – 30x + 13**

Use Calculator to find two zeros. Answer: {1,2-3i,2+3i}

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**Zeros of Polynomial Functions**

Complex Numbers (a+bi) Imaginary Numbers (+bi) REAL number system Rational Numbers Irrational Numbers

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**General shapes of graphs with a positive leading coefficient.**

Degree 1 Degree 2 Degree 3 1 zero 2 zeros 3 zeros Degree 5 Degree 4 5 zeros 4 zeros Remember, zeros are just x-intercepts.

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**Finding a Polynomial Function with Given Zeros**

EXAMPLE 1: Find a 3rd degree polynomial function f(x) with real coefficients that has -3 and i as zeros and such that f(1) =8. f(x)= an (x-c1)(x-c2)(x-c3) Now substitute in the zeros with what you know. Do not forget about the conjugate pairs. f(x) = an(x+3)(x+i)(x-i) Multiply the polynomial out.

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**f(x)= an(x3 + 3x2 + x + 3) f(1)= an[(1)3 + 3(1)2 + 1 +3] = 8an 8 = 8an**

Polynomial Equation is f(x) = (x+3) (x2+1) or x3 + 3x2 +x + 3 YOU TRY: Find a 3rd degree polynomial function f(x) with real coefficients that has 4 and 2i as zeros and such that f(-1) =-50. Answer: f(x) = 2x3 – 8x2 +8x -32

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Summary: Describe how to find the possible rational zeros of a polynomial function.

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