Zeros of Polynomial Functions

[nycmathguy]

Section 2.5
10 & 12

Question 10

f(x) = x^4 - 3xp

f(x) = x(x^3 - 3)

Set each factor to 0 and solve for x.

x = 0

x^3 - 3 = 0

x^3 = 3

Take cube root on both sides.

x = 3^(1/3)

This function has 2 zeros.

Question 12

f(x) = x^3 - x^6

f(x) = x^3(1 - x^2)

Set each factor to 0 and solve for x.

x^3 = 0

Let cr = cube root.

cr{x^3} = cr{0}

x = 0

1 - x^2 = 0

-x^2 = -1

x^2 = -1/-1

x^2 = 1

sqrt{x^2} = sqrt{1}

x = 1

This function also has 2 zeros.

You say?

 

[MathLover1]

Question 10
given f(x) = x^4 - 3x is degree 4, so it has 4 zeros
one is x=0
x = 3^(1/3) -> it's cube root, so this function has 3 zeros-> one real and two complex zeros

Question 12
f(x) = x^3 - x^6
when you factor it, you get
x^3 - x^6=x^3(1 - x^3)

so zeros are
cr{x^3} = cr{0} which is x = 0 (0 is a triple root)
and
(1 - x^3)=0
1 =x^3
cr{1}=> one real zero and two complex zeros

so,
Real solutions:

Complex solutions

 

[nycmathguy]

Question 10
given f(x) = x^4 - 3x is degree 4, so it has 4 zeros
one is x=0
x = 3^(1/3) -> it's cube root, so this function has 3 zeros-> one real and two complex zeros

Question 12
f(x) = x^3 - x^6
when you factor it, you get
x^3 - x^6=x^3(1 - x^3)

so zeros are
cr{x^3} = cr{0} which is x = 0 (0 is a triple root)
and
(1 - x^3)=0
1 =x^3
cr{1}=> one real zero and two complex zeros

so,
Real solutions:

Complex solutions

In other words, I am wrong yet again. Right? Why is it necessary to find the complex solutions, too?

 

[MathLover1]

right, sqrt{1} is not same as cr{1}

it's important to find all solutions unless question ask you to find real solutions only

 

[nycmathguy]

right, sqrt{1} is not same as cr{1}

it's important to find all solutions unless question ask you to find real solutions only

In that case, I will stop right here. Moving forward, I will do my best to actually read the lecture notes presented per section and try to make sense of what Ron Larson is saying. What does this mean? It means that I am going to SLOW DOWN my review of precalculus. I am anxiously trying to run through the textbook rushing my way to Calculus 1. I will not post new questions from Section 2.5 until some reading of the lecture notes is accomplished.