1. ## polynomial

hi, i really wanna ask a question is there any easier way to factor this polynomial 6x^4 + x^3 - 8x ^2 -x + 2 ,
and i have one more question, when do use the integral zero theorm P (b)=0and the factor theorm P ( b/a ) = 0. i just wanna know the difference , and when to use which. thanks

2. Originally Posted by lickman
hi, i really wanna ask a question is there any easier way to factor this polynomial 6x^4 + x^3 - 8x ^2 -x + 2 ,
and i have one more question, when do use the integral zero theorm P (b)=0and the factor theorm P ( b/a ) = 0. i just wanna know the difference , and when to use which. thanks
$
f(x) = 6x^4+x^3-8x^2-x+2$

If you put x=1 and x=-1 in the function , it becomes 0. so the factors (x-1) and (x+1) are the factors of f(x).

Now divide f(x) with the product $=(x-1)(x+1) = x^2 - 1$, we get a quadratic,

$6x^2+x-2$

$6x^2+x-2= (2x-1)(3x+2)$

so, factors are

$f(x) = 6x^4+x^3-8x^2-x+2=(x-1)(x+1)(2x-1)(3x+2)$

3. thanks alot, dividng is such a long procedure, lol. i have one more question about the P(B) = 0 and P(B/A) = 0, when do u include a and when do u only use B ? i often get confuse on trying to find the root with those

4. Originally Posted by lickman
thanks alot, dividng is such a long procedure, lol. i have one more question about the P(B) = 0 and P(B/A) = 0, when do u include a and when do u only use B ? i often get confuse on trying to find the root with those
There might be times when substituting integer values into the polynomial won't give 0 as a result. If this is the case (and you can probably tell by the constant term, if it's a fraction) then you might need to substitute fractions.