Arb.dha
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Anon???
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Use factor theorem. If you sub in the value of x into p(x) and you get 0, it’s a factor
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Arb.dha
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(Original post by Anon???)
Use factor theorem. If you sub in the value of x into p(x) and you get 0, factor
I tried that. The following questions ask for that and it said they're right. it said this isn't I'm not sure why.
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DFranklin
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Post the actual question.
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Arb.dha
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(Original post by DFranklin)
Post the actual question.
I've edited now
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0le
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(Original post by Arb.dha)
I've edited now
Your attachment just says "image not found".

Secondly, look at your original post:

(Original post by Arb.dha)
the cubic function p(x)=2x^3 x^2 -13x 6
You have not stated the signs on some of the terms. This therefore has several possibilities:

2x^3 + x^2 - 13x + 6
2x^3 + x^2 - 13x - 6
2x^3 - x^2 - 13x - 6
2x^3 - x^2 - 13x + 6

There are some tricks to factoring polynomials but a lot of the time it is simply just a case of trial and error. So just plug in some numbers and try.

The process of what you are really doing is as follows. You want to get something of the form:
2x^3 + x^2 - 13x + 6  = (x + a)(Ax^2 + Bx + C) = 0

Note that we can write it like this because I presume the question implies that it can be reduced to a simpler form. This is not always the case - sometimes you cannot find any factors!

Here, a can be any rational number from the list you've given and this is where trial and error is needed. You need to figure out if a = 1 or if a = -2 etc.

Note that A and B are the coefficients and C is a real number. You can use polynomial division to find the second polynomial, (Ax^2 + Bx + C) i.e.:

(Ax^2 + Bx + C) = \frac{2x^3 + x^2 - 13x + 6}{x + a}:

If you need to determine x, then you need to use something called the "zero product property of real numbers". This is a fancy way of saying the following. If you have two real numbers g and h, if their product is gh = 0, then either g or h or both must be zero.

In your original polynomial, you have:
(x + a)(Ax^2 + Bx + C) = 0

Here (x + a) is one number and the other number is (Ax^2 + Bx + C). Therefore you can write (x + a) = 0 and (Ax^2 + Bx + C) = 0 and solve accordingly after you have figured out all the factors.
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Plücker
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(Original post by Arb.dha)
Can anyone help?
I have the following question:
the cubic function p(x)=2x^3 x^2 -13x 6
which of the following are possible factors of p(x)?
x /- 1
x /-2
x /-3
x /-4
x /-6
x /-0.5
x /-1.5
I have tried to factorise it. that doesn't get the right answer. You can pick multiple ones but I just don't understand how to get to the answer. Can anyone help?
Edit: attached image
Attachment 1020387
It's a stupid question. Especially when you're then asked to find the factors and you get it right.

What's their definition of a "possible factor"? It's either a factor or it isn't.

I know what the intention is but having seen a student using this online course I know that there's a problem with this question.
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