Question

Which is a possible number of real roots for a cubic function? Select all that apply.

0
1
2
3
4

what even is this

Answer 1

is it all of them

Answer 2

what is the highest power on the variable in a cubic function?

Answer 3

no idea

Answer 4

do you know what a cubic equation is?

Answer 5

no

Answer 6

google

Answer 7

So, any polynomial can have two types of roots:
-Real roots, composed of real numbers
-Imaginary (non-real) roots, with some imaginary component to them.

So, depending on the polynomial we're dealing with, our roots could be imaginary, real or a mixture of both.

When you think about a 'CUBIC' polynomial, how many roots overall do you think it will have @dirtydan667 ?

Answer 8

three

Answer 9

correct

Answer 10

so i would click all execpt 4 beacasue it says select all that apply

Answer 11

yes

Answer 12

what you typing ....lol

Answer 13

if your explaining something i wanna see

Answer 14

Yes! A cubic polynomial will be of the form:

\[ax ^{3} + b x^{2} + cx + d\]

where a, b, c and d are some kind of real numbers.

As you said, this will always have three roots. These roots could be all real roots, meaning 3 is a possible answer to our equation. For any imaginary (non-real) roots, they MUST come in what are called 'conjugate pairs'. That means, of our cubic has an imaginary root
\[a + bi\]
where 'i' is the imaginary part, then it must also have another imaginary root
\[a - bi\]
hence why we call them 'pairs'!!

So, based on this, what's the other possible number of real roots for a cubic @dirtydan667 ?

Answer 15

infinate ?

Answer 16

anything but 4 ?

Answer 17

(0,1,2,3)

Answer 18

So, we know we have 3 roots....

We dealt with the first case, where we know all 3 of these roots will be real roots. So, 3 is a possible answer.

We're now dealing with the second case, whereby we know that some of our roots will be imaginary. If we have imaginary roots, they must come in pairs, so we can't have an odd number of imaginary roots.

This means that we can't have a cubic with 3 imaginary roots, as one of these would be unpaired. Similarly, we can't have a cubic with 1 imaginary root, as this wouldn't have a pair either. But, we can have one with 2 imaginary roots (a conjugate pair!).

So, if we have 3 roots and 2 of them are imaginary, that means that the third one must be real....so 1 real root is the other possible answer!! Are you able to follow that @dirtydan667 ? :)

Answer 19

Are you able to understand why this is the case @dirtydan667 ?