Question

f(x) = x3 + 3x2 – 10x – 24. I need to identify the correct factors of f(x).

These are suggestions:

(x – 1)
(x + 2)
(x – 3)
(x + 4)
(x + 6)
(x – 12)

List the correct factors. Then justify your selections with complete sentences.

Answer 1

@phi @satellite73 @Hero

Answer 2

x - 1 is a factor of f(x) if f(1) = 0
x + 2 is a factor of f(x) if f(-2) = 0
x - 3 is a factor of f(x) if f(3) = 0

See a pattern?

Answer 3

Ohh

So i solve for them until one get me an answer of 0?

Answer 4

It will probably be more than one

Answer 5

Now to check if x - 1 is a factor:

f(1) = (1)^3 + 3(1)^2 - 10(1) - 24 = ?

Answer 6

Yes, when you find all of the factors that equal zero, then you will have all of them.

Answer 7

Wouldn't it be (f-1) = (-1)^3 + 3(-1)^2 - 10(-1) - 24 = ?

Answer 8

No, don't confuse yourself

To find the factors, this is what you do:

x - 1 = 0
x = 1

Answer 9

Which means since x = 1 is a possible factor, we have to substitute 1 in place of x for f(x). So find f(1). If it equals zero, then you have found one of the factors of f(x). If it doesn't then move on to the next possible factor.

Answer 10

So every factor would be positive?

Answer 11

Think back to the zero product property in order to understand

Answer 12

Thank you :)

Answer 13

I have another question, think you could help?

• 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
• 2 is a zero of g(x) because g(2) = 0.
• 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.

Correct the reasoning of any inaccurate statements. Make sure you reference any theorems that support your justifications.

Answer 14

This site was designed so that one question per thread would be posted. Notice that in the blue box at the top of this thread is reserved for questions. The white boxes are reserved for responses.

Answer 15

In other words, if you have another question, the correct course of action is to close this question, then post a new question.

Answer 16

If I start a new thread and tag you could you help? I figured that since you were already here I might as well post it...

Answer 17

If you have another question, close this one, and post the other question. If you want to tag me or someone else, that's fine.