Chikis Posted April 21, 2016 Share Posted April 21, 2016 I want to use the factor theorem to find the factors [math]2x^3+x^2-13x+6[/math] How do I do that? Link to comment Share on other sites More sharing options...

ajb Posted April 21, 2016 Share Posted April 21, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 (edited) It says that [math]x-r[/math] is a factor of a polynomial [math]f( x)[/math] if [math]f( r)=0[/math] Edited April 21, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 21, 2016 Share Posted April 21, 2016 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). Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 (edited) What is 'r' and what is [math]f( r)[/math] Edited April 21, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 21, 2016 Share Posted April 21, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 What does the r and f stand for? Or what does [math]f( r)[/math] stand for? Link to comment Share on other sites More sharing options...

ajb Posted April 21, 2016 Share Posted April 21, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 (edited) When [math]x=2[/math] [math]\rightarrow[/math] [math]2(2)^3+(2)^2-13(2)+6[/math] [math]\rightarrow[/math] [math]16+4-26+6=0[/math] [math]\rightarrow[/math][math]26-26=0[/math] [math]\therefore[/math] [math]x-2[/math] is a factor of [math]2x^3+x^2-13x+6[/math] When [math]x=-3[/math] [math]2(-3)^3+(-3)^2-13(-3)+6[/math] [math]=2(-27)+9+39+6[/math] [math]=-54+48+6[/math] [math]=-54+54=0[/math] [math]\therefore[/math][math]x+3[/math] is a factor of [math]2x^3+x^2-13x+6[/math] But what is the easiest method for finding the values of x? Or would one be always doing trial and error method? Edited April 22, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 22, 2016 Share Posted April 22, 2016 You should try factors of 6, in this case. Can you see why? Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 (edited) 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 April 22, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 22, 2016 Share Posted April 22, 2016 (edited) 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 April 22, 2016 by ajb Link to comment Share on other sites More sharing options...

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