Solving Polynomial Equations

Section 4.5 beginning on page 190

Solving By FactoringWe already know how the zero product property allows us to solve quadratic equations, this property also allows us to solve factored polynomial equations [we learned how to factor polynomial expressions in the previous section].

Example 1: Solve

2 𝑥=0 𝑥−3=0𝑥=0 𝑥=3

** Because the factor occurred twice, the root has a multiplicity of 2.

** When a factor of a function is raised to an odd power, the graph of the function crosses the x-axis at

** When a factor of a function is raised to an even power, the graph of the function touches the x-axis at

Finding Zeros and GraphingExample 2: Find the zeros of . Then sketch a graph of the function.

𝑓 (𝑥 )=−2 𝑥4+16 𝑥2−320=−2(𝑥4−8 𝑥2+16)

0=−2(𝑥2−4)(𝑥2−4 )

0=−2(𝑥+2)(𝑥−2)(𝑥+2)(𝑥−2)0=−2(𝑥+2)2(𝑥−2)2

The zeros are 2 and -2. Both factors are raised to an even power, so the graph will touch the x-axis at those points.

This is an even degree polynomial with a negative leading coefficient so the ends both go down.

The y-intercept is -32

𝑢=𝑥2

𝑢2−8𝑢+16(𝑢−4)(𝑢−4 )

The Rational Root Theorem

This theorem gives a starting point for factoring some higher degree polynomials.

We will often list the possible factors, and then test them to find one factor and go from there.

Using the Rational Root TheoremExample 3: Find all real solutions of

Factors of 20: 1,2,4,5,10,20Factors of 1: 1

List the possible solutions:

Test the possible solutions using synthetic division until you find one that works.

𝑥=±11, ±21,±41,±51, ±101, ±201

Continued….Example 3: Find all real solutions of

Lastly, factor completely using the result of the synthetic division.

𝑥+1=0 𝑥−5=0𝑥=−1 𝑥=5

𝑥−4=0𝑥=4

The Irrational Conjugate Theorem

This is simply saying that if is a solution to function, then is also a solution to the function.

Using Zeros to Write a PolynomialExample 5: Write a polynomial function of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 3 and

** Because of the irrational conjugate theorem, we know that is also a zero of this function.

𝑓 (𝑥 )=¿(𝑥−3)[𝑥− (2+√5 )][𝑥− (2−√5 )] Step 1: Write in factored form.

Step 2: Regroup the terms

(notice how this is a special product)

Step 3: Multiply everything out and simplify

𝑓 (𝑥 )=(𝑥−3 ) [𝑥−2−√5 ] [𝑥−2+√5]𝑓 (𝑥 )=(𝑥−3 ) [(𝑥−2)−√5 ] [(𝑥−2)+√5 ]𝑓 (𝑥 )=(𝑥−3 ) [ (𝑥−2 )2−5 ]𝑓 (𝑥 )=(𝑥−3 )¿𝑓 (𝑥 )=(𝑥−3 ) [𝑥2−4 𝑥−1 ]𝑓 (𝑥 )=𝑥3−4 𝑥2−𝑥−3 𝑥2+12 𝑥+3𝑓 (𝑥 )=𝑥3−7 𝑥2+11𝑥+3

Monitoring ProgressSolve the equation:

1) 2)

Monitoring ProgressFind the zeros of the function. Then sketch a graph of the function.

3) 4)

Monitoring Progress5) Find all real solutions.

7) Write a polynomial function of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 4 and