# Factor theorem

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I want to use the factor theorem to find the factors $2x^3+x^2-13x+6$ How do I do that?

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What does the factor theorem say?

From there the usual thing to do is try to make an educated guess at one of the possible roots. Think about this carefully. Whatever the factorisation of this is, the parts that have nothing to do with x must multiply up to 6. This will give you some ideas of what to try.

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It says that $x-r$ is a factor of a polynomial $f( x)$ if $f( r)=0$

Edited by Chikis
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Right, so you need to find an 'r' such that f[r] =0 and then you know you can write you polynomial as (x-r)(ax^2 + bx +c).

So try a couple of values for 'r' and see if you can find one root. From there you maybe able to factor the quadratic or apply the factor theorem again... or it may not factor (over the reals).

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What is 'r' and what is $f( r)$

Edited by Chikis
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You need to find some values of x =r so that the polynomial is zero. This gives you one of the roots. You can just throw random numbers into the polynomial or try to be a bit more clever than that. I gave you a hint earlier for some sensiable guesses.

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What does the r and f stand for? Or what does $f( r)$ stand for?

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f is the polynomial in question, we often write this as f[x] for the variables x. f[r] I understand as evaluating the polynomial at x =r.

This is all quite standard notation.

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When $x=2$

$\rightarrow$

$2(2)^3+(2)^2-13(2)+6$

$\rightarrow$

$16+4-26+6=0$

$\rightarrow$$26-26=0$

$\therefore$ $x-2$ is a factor of

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

When $x=-3$

$2(-3)^3+(-3)^2-13(-3)+6$

$=2(-27)+9+39+6$

$=-54+48+6$

$=-54+54=0$

$\therefore$$x+3$ is a factor of $2x^3+x^2-13x+6$ But what is the easiest method for finding the values of x? Or would one be always doing trial and error method?

Edited by Chikis
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You should try factors of 6, in this case. Can you see why?

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No I can't. Could you explain why? Or should I take it that the rule is to use the factors of the constant term.

Edited by Chikis
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You are looking to write your order three polynomial ax^3 + bx^2 + cx + d as something like (x-A)(B x^2 + C x +D) and then maybe factorise the quadratic also. Notice the just by expanding this expression you need d = AD. Thus A must be a factor of d (assuming all integers etc.)

This will also work for higher order. It allows you to make some initial guesses without being totally random.

Edited by ajb

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