f(x)=x^3-5x^2-14x
f(x)=x^3+4x^2+9x+36
f(x)=x^4+x^3-2x^2+4x-24
i really need the helpp!
thanks in advance!
simply factor out an x and you obtain a quadratic to factorize.
factor by grouping. there are common terms among the first two terms and the last two terms of the original cubicf(x)=x^3+4x^2+9x+36
by the rational roots theorem (look it up), if there are rational roots to this equation, they will be from the factors of 24 (namely, $\displaystyle \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \mbox{ and } \pm 24$). start with the ones and plug them in and see if you get zero. if you obtain 0 for one, then it means that is a root. you can repeat this process with the others. or you may prefer to use polynomial long division or synthetic division to break down the polynomial in the following way.f(x)=x^4+x^3-2x^2+4x-24
let $\displaystyle a$ be one of the roots you found using the above method. so $\displaystyle a$ is a root. by the factor theorem, $\displaystyle (x - a)$ is a factor of the equation. so use long (or synthetic) division to compute $\displaystyle (x^4 + x^3 - 2x^2 + 4x - 24) \div (x - a)$
you will obtain a cubic as the result. for which you can repeat the process (or perhaps factoring by grouping can work). this way you keep breaking it down until you get to a quadratic, where factoring (i hope) is easy for you.
try them, good luck. if you don't understand something, say so