Week 6: Zeros of Functions

This week the topic was Zeros of Functions. On Day three and Day four, we covered zeros of Functions.

On Day three we looked at Zeros and Factors of Polynomial Functions. 

We looked at 

f(x)=d(x)q(x)+r(x)

if r(x)=0, then d(x) is a factor of f(x)

We also looked at

f(x)/x-c × r(x)=f(c)

We tried to look for factors and zeros.

The steps to do so:

1. Complete the factorization

2. List the zeros

The difference between a factor and a zero is

ex:

Factor | Zero

x-7     7

x+9   -9

4x-1   1/4

x-14    14

Also a reminder from Algebra 2 is:

i = √-1

i^2 = -1

a+bi = complex #

a-bi = conjugate

(a+bi)(a-bi) FOIL

Conjugate o’s always occur in conjugate pairs

ex: 1-i <---> 1+i

2i <---> -2i

So we used synthetic division in this type of problem. You would do the exact same thing from the previous lesson but the remainder must be zero. If the remainder is not zero, then the equation does not work. It would not have a zero or a factor.

You would do the same steps with a remainder of zero and then after that. You would turn the number given or the original factor and turn that into a factor: ex (x-2). Then the equation provided from the end. You would try to factor that to get the other factors. Sometimes you would use the Quadratic formula to solve it. Then it would look like for example: (x-2)(x-1)(x+9). that would be the factor and the zeros would be x=-9, 1, 2. That would be the answer. 

HINT: the first degree in the original equation tells you how many zeros there will be in a problem. Sometimes there would be imaginary numbers to make the remaining zeros. For example: x^2-4. There will be 2 zeros in this equation.

Below is two examples of how it would work.

In the second problem, there are imaginary numbers, which appear when you plug it into the quadratic formula to find out how to solve an equation. That is how you find the zeros and factors of a polynomial function.

On day four, we also look at Real Zeros of Polynomial Functions.

In this one we had to so which zeros were real inside a polynomial function. To do this, we had to do 

P= Factors of constant

s   Factors of leading coefficient

The rules on how to find is 

1. Find all possible zeros P/S

2. Test zeros using remainder thereom

HINT: use graph to see possible zeros

This was done in Algebra 2 in the previous year and to begin the problem: a Polynomial will be given to you.

There will be many zeros listed and you are trying to find the real ones and when a factor is not given. Below is an example of how to do it. 

So first you would look at the constant which is the number without x or with a variable. You would find all the possibilities on how to make that number. 

For example from above. -2: to make -2 you could use ±1 and ±2. You would put that on the line for P.

Next you would look at the first coefficient, which is above 3x^4.

You would see how many possibilites would go into 3, which is ±1 and ±3. You would put that on the line for S.

Just like above you see all the it has ±1 and ±2 over ±1 and ±3.

You then times all the possibilites by each other.

For example: 

±1/±1 is one or negative one. Then ±1/±3 is 1/3 or -1/3.

±2/±3 is 2/3 or -2/3

±2/±1  is ±2

So x= -1, 1, 1/3, -1/3, -2, 2, -2/3, 2/3

You then put those into x to figure out After you find one that is equal to 0.

After you find it equal to 0 you put the equation given into Synthetic Division with the zero that works. You then try more zeros and put it into the simplified equation until you have all the zeros needed. Just as this equation needs 4 exact zeros. 

That is how to find the zeros of a function.

See You Next Week!