Does anybody know the p over q method?

Question

anonymous · Answer

for what?  finding "possible rational zeros of a polynomial"?

anonymous · Answer

or something else?

anonymous · Answer

i think, it says use the p over q method and synthetic division to factor the polynomial p(x) . then solve p(x)=0.

example 1) p(x) = $$x ^{3}+4x ^{2}+x-6$$

anonymous · Answer

ok you want possible rational exponents.  that is what you are after. in this case easy

anonymous · Answer

numerator has to divide 6 and denominator has to divide 1 so only possibilities are 
$$\pm1,\pm2\pm3\pm6$$

anonymous · Answer

1 is the easiest to try and it works with your eyeballs

anonymous · Answer

$$p(1)=1+4+1-6=0$$

anonymous · Answer

so you know this factors as 
$$p(x)=x^3+4x^2+x-6=(x-1)\times q(x)$$

anonymous · Answer

and you can find 
$$q(x)$$ either by polynomial division, synthetic division, or thinking.

anonymous · Answer

positive 1 works right?

anonymous · Answer

yes easy.  add up the coefficients, see that you get 0

anonymous · Answer

okay when i did synthetic for p(1) i got $$1, 4, 1, -6 $$ after i multiplied by one i ended up getting x^2 +5x +6, because i had a remainder of 0, so that factored into (x+3)(x+2) did i do something wrong?