Help with factoring inequalities with 4 terms

[wind_surfer]

Hello, I have started learning about inequalities and so far I have just been graphing them in order to find where the x-intercepts are and from there stating the solution. However now I am supposed to start figuring them out algebraically, I can do the ones with 3 terms easily but I am having difficulty understanding how to factor with 4 terms.
For example: x^3-6x^2+5x+12>0
I know it factors to (x+1)(x-3)(x-4) based on the graph it produces, but I have no idea how to get there.
Can someone explain the strategy? Or at least give me a link to a website that explains it clearly?
Any help would be appreciated.

Thankyou

 

[Deleted member 4993]

Hello, I have started learning about inequalities and so far I have just been graphing them in order to find where the x-intercepts are and from there stating the solution. However now I am supposed to start figuring them out algebraically, I can do the ones with 3 terms easily but I am having difficulty understanding how to factor with 4 terms.
For example: x^3-6x^2+5x+12>0
I know it factors to (x+1)(x-3)(x-4) based on the graph it produces, but I have no idea how to get there.
Can someone explain the strategy? Or at least give me a link to a website that explains it clearly?
Any help would be appreciated.

Thankyou

Do you know rational root theorem?

If yes - use it.

If not then - for a starter go to:

http://www.purplemath.com/modules/rtnlroot.htm

There is a very complicated method of finding x-intercepts of cubics (Cardan's method) - but I suppose that is not at your level.

 

[HallsofIvy]

I presume you know that x- a is a factor of a polynomial if and only if x= a makes the polynomial equal to 0. So factoring P(x) is the same as solving P(x)= 0. That is, in general, very difficult but if you can make a good "guess" at what the roots might be you can try just evaluating the polynomial at those various "guesses" and see if they make the polynomial equal to 0.

And a good way of starting is the "rational root theorem" that Subhotosh Khan mentions.
"If a rational number, m/n, satisfies the polynomial equation with integer coefficients, \(\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0\) then the numerator, m, evenly divides the "constant term", \(\displaystyle a_0\), and the denominator, n, evenly divides the "leading coefficient", \(\displaystyle a_n\)."

Notice that does NOT say that such an equation must have rational roots and so does not say that this will give the roots- only that if there are rational roots, they must be of this form so this is what we should try. The polynomial you give, \(\displaystyle x^3-6x^2+5x+12\) has "leading coefficient" 1 and the only integers that evenly divide 1 are 1 and -1. The "constant term" is 12 and the only integers that evenly divide 12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12. That tells us that if the equation \(\displaystyle x^3-6x^2+5x+12= 0\) has any rational roots they must be among 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12. Trying those in the equation, we see that \(\displaystyle (-1)^3- 6(-1)^2+ 5(-1)+ 12= -1- 6- 5+ 12= 0\), \(\displaystyle (3)^3- 6(3)^2+ 5(3)+ 12= 27- 54+ 15+ 12= 0\) and that \(\displaystyle (4)^3- 6(4)^2+ 5(4)+ 12= 64- 96+ 20+ 12= 0\). Thus x= -1, x= 3, and x= 4 are roots of the equation so x-(-1)= x+ 1, x- 3, and x- 4 are factors of the polynomial.