How Do You Factorise Polynomial Expressions?

  • Thread starter Thread starter roxas
  • Start date Start date
AI Thread Summary
To factorise the polynomial expression x^3 - 2x^2 - 5x + 6, start by finding a zero, which can be guessed from the factors of 6, such as -1, 1, 2, -2, 3, -3, 6, or -6. Once a zero is identified, perform Euclidean division of the polynomial by (x - zero) to simplify it into a second-degree expression. Some participants expressed frustration with the original poster's lack of effort in solving similar problems independently. Despite this, a step-by-step approach was suggested to aid understanding. The discussion highlights the importance of engaging with the material rather than seeking direct answers.
roxas
Messages
21
Reaction score
0
i know i totally sux so can anyone help me? thx

factorise x^3 - 2x^2 - 5x + 6
 
Physics news on Phys.org
Find a zero for your expression.
 
i am too retarded:smile: , can u show me a step by step please
 
arildno said:
Find a zero for your expression.
Yep and you should start by "guessing a number" that is either (-1,1,2,-2,3,-3,6,-6). Why these numbers ? Well 6 can be divided by these numbers. Let's say that number is 6. Then perform an Euclidian devision in which you divide the given expression by (x - 6). What you get is an expression of second degree in x times (x-6) equals the given expression. This should get you started

marlon
 
Last edited:
Don't bother to help OP, marlon.
In another thread, he has refused even to do a single thing on his own with the expression 18t^2-2.

He is just after having his homework done for him.
 
arildno said:
Don't bother to help OP, marlon.
In another thread, he has refused even to do a single thing on his own with the expression 18t^2-2.

He is just after having his homework done for him.
Opps did not know that. Thanks for the advice, besides enough info has been given.

regards
marlon
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top