Q: how can you figure out if the sign of f on the interval -1 < x < 4 is positive or negative?

I guess you're talking about 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 4)², which has the zeros 𝑥 = 1 and 𝑥 = 4 In expanded form we have 𝑓(𝑥) = 𝑥³ − 9𝑥² + 24𝑥 − 16 Because the leading coefficient is positive and the degree of 𝑓 is odd, 𝑓 will be negative for 𝑥 < 1 Because the multiplicity of 𝑥 = 1 is odd, 𝑓 will be positive over the interval 1 < 𝑥 < 4

Q: What if there is a problem like (x-1)^3 (x+2)^2 will the multiplicity be the addition of 3 and 2 or the highest exponent will be the multiplicity?

A polynomial doesn't have a multiplicity, only its roots do. The roots of your polynomial are 1 and -2. 1 has multiplicity 3, and -2 has multiplicity 2.

Q: Using multiplity how can you find number of real zeros on a graph

So first you need the degree of the polynomial, or in other words the highest power a variable has. So if the leading term has an x^4 that means at most there can be 4 0s. There can be less as well, which is what multiplicity helps us determine. If a term has multiplicity more than one, it "takes away" for lack of a better term, one or more of the 0s. So for instance (x-1)(x+1)(x-2)(x+2) will have four zeros and each binomial term has a multiplicity of 1 Now, if you make one of them have a multiplicity of 2 that takes away one of the zeroes. so (x-1)(x-1)(x+2)(x-2), here there are two (x-1) terms so it has multiplicity 2, this means there is one less zero. So now there are only three zeroes at 1, 2 and -2. ALSO if a term has an even multiplicity it means it touches the x axis rather than crosses it. Let me know if that didn't help.

Question 1

Why does the graph only touch the x axis at a zero of even multiplicity?

Accepted Answer

I've been thinking about this for a while and here's what I've come up with.
Let's say, for example, that f(x) = ( x - 4 ) ( x - 1 )^2.
( x - 4 ) is a root of odd multiplicity.
Notice that when x < 4, ( x - 4 ) is negative,
but when x > 4, ( x - 4 ) is positive.
So, depending on the value of x, the sign of ( x - 4 ) changes, which in turn changes the sign of f(x).
But also notice that for roots of even multiplicity [ ( x - 1 ) in this example], it doesn't matter what value of x is chosen.  Once raised to their EVEN power, they will always be positive, so will not be able to change the sign of f(x).
So, if f(x) is negative as it approaches a zero of EVEN multiplicity, then f(x) will remain negative after it passes that zero (and likewise if f(x) was positive, it would remain positive).  In other words, it would just touch the x-axis and then have to "bounce" away in the same (positive or negative) direction.
But if f(x) is negative (or positive) as it approaches a zero of ODD multiplicity, then f(x) will change sign --- in other words, the graph will cross through the x-axis instead of bouncing back.
I hope this has been helpful and hasn't ended up confusing you!

Question 2

in the answer of the challenge question 8 how can there be 2 real roots . in total there are 3 roots as we see in the equation . but in the answer there are 2 real roots which will tell that there is only 1 imaginary root which does not exists. please help me . thanks in advance!!

Accepted Answer

There is no imaginary root.  Sometimes, roots turn out to be the same (see discussion above on "Zeroes & Multiplicity").  That is what is happening in this equation.  So, the equation degrades to having only 2 roots.  
If you factor the polynomial, you get factors of:  -X (X - 2) (X - 2).  You can see, 2 of the factors are identical.
If you use these to solve for f(x) = 0, they create only 2 points:  (0,0) and (2,0) because we have 2 identical factors that both create X=2.
Hope this helps.

Question 3

For problem Check Your Understanding 6), if its "6", then why is it odd, not even?

Accepted Answer

The question asks about the multiplicity of the root, not whether the root itself is odd or even.
At a root of odd multiplicity, the graph will cross through the X-axis.
At a root of even multiplicity, the graph will bounce off the X-axis and not go through it.

Question 4

In challenge problem 8, I don't know understand how we get the general shape of the graph, as in how do we know when it continues in the positive or negative direction. So for example, from left to right, how do we know that the graph is going to be generally decreasing?

Accepted Answer

You don't have to know this to solve the problem. You can find the correct answer just by thinking about the zeros, and how the graph behaves around them (does it touch the x-axis or cross it). You can click on "I need help!" to see the solution.

If you want to know how to determine the direction of the graph, check out the next tutorial:

https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-end-behavior/a/end-behavior-of-polynomials

Question 5

Why is Zeros of polynomials & their graphs important in the real world, when am i ever going to use this?

Accepted Answer

It depends on the job that you want to have when you are older. School is meant to prepare students for any career path, including those that have to do with math. You might think now that you don't want a career with math, but you never know if you might decide to change your aspirations. When my mother was a child she hated math and thought it had no use, though later in life she actually went into a career that required her to have taken high math classes. You might use it later on! I'm grateful enough that I even have the opportunity to have such a nice education compared to developing countries where most citizens never make it to college.

Question 6

How would I solve f(x)=(2x-1)(x-5)?

Accepted Answer

You have a function with infinite solutions.  It can be solved by using any input value for "x" and calculating "y".

What where you asked to find?  If you were asked to find the zeroes, they are the X-intercepts.  Use y=0 and find x.
(2x-1)(x-5) = 0
The polynomial is already in factored form.  So, use the zero product rule to split the factors apart
2x-1=0 and x-5=0
Solve these and you have the zeros.
Hope this is what you meant by "sovle".

Question 7

how can you figure out if the sign of f on the interval -1 < x < 4 is positive or negative?

Accepted Answer

I guess you're talking about 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 4)²,
which has the zeros 𝑥 = 1 and 𝑥 = 4

In expanded form we have 𝑓(𝑥) = 𝑥³ − 9𝑥² + 24𝑥 − 16

Because the leading coefficient is positive and the degree of 𝑓 is odd,
𝑓 will be negative for 𝑥 < 1

Because the multiplicity of 𝑥 = 1 is odd,
𝑓 will be positive over the interval 1 < 𝑥 < 4

Question 8

What if there is a problem like (x-1)^3 (x+2)^2 will the multiplicity be the addition of 3 and 2 or the highest exponent will be the multiplicity?

Accepted Answer

A polynomial doesn't have a multiplicity, only its roots do. The roots of your polynomial are 1 and -2. 1 has multiplicity 3, and -2 has multiplicity 2.

Question 9

Using multiplity how can you find number of real zeros on a graph

Accepted Answer

So first you need the degree of the polynomial, or in other words the highest power a variable has.  So if the leading term has an x^4 that means at most there can be 4 0s.  There can be less as well, which is what multiplicity helps us determine.  If a term has multiplicity more than one, it "takes away" for lack of a better term, one or more of the 0s.

So for instance (x-1)(x+1)(x-2)(x+2) will have four zeros and each binomial term has a multiplicity of 1  Now, if you make one of them have a multiplicity of 2 that takes away one of the zeroes.  so (x-1)(x-1)(x+2)(x-2), here there are two (x-1) terms so it has multiplicity 2, this means there is one less zero.  So now there are only three zeroes at 1, 2 and -2.  ALSO if a term has an even multiplicity it means it touches the x axis rather than crosses it.

Let me know if that didn't help.

Course: Get ready for AP® Calculus > Unit 5

Zeros of polynomials & their graphs

What you will learn in this lesson

Fundamental connections for polynomial functions

Check your understanding

Zeros and multiplicity

Check your understanding

The graphical connection

Check your understanding

Challenge problem

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