How to find number of rational zeros?

Question

how many rational zeros are in: f(x) = x^3 + 2x^2 - 11x - 12
and in: g(x) = x^3 - 9x^2 + 15x + 25

i can't figure out how to factor this? how do you do this?
Thanks!

Answer 1

for f(x)
use the factorisation theorem,put f(x)=0
since the last value is -12 there must be one negative number...so try using -1
f(-1)=-1+2+11-12=0
so x+1 is a factor,now divide f(x) by x+1
you will get x^2+x+-12
whose factors are -4,+3
so the rational factors are x=-1,-4,+3
there is another hint that you can always use
both complex roots and irrational roots always occur in conjugate pairs..meaning if one root is +a their other root will be -a in the case of irrational roots
in case of complex roots, it will be a+ib and a-ib
so you could also use that as a hint for any problem of that type
for g(x)
x=-1 is a root
then divide g(x) by x+1
then u'll get
x^2-10x+25
whose roots are 5,5
so your zeroes are -1,5

Answer 2

f(x) = x^3 + 2x^2 - 11x - 12
has 3 rational 0's. The factors are
(x - 3)(x + 4)(x + 1)
Set the function equal to 0 and try the factors of - 12 one at a time. When you find one that works you can choose to divide the polynomial by the term or continue trying untried factors.

g(x) = x^3 - 9x^2 + 15x + 25
has 2 rational 0's. The factors are
- 1 - 9 = - 10
- 10*-1 + 15 = 25
25*-1 + 25 = 0
so - 1 is a root and (x + 1) is a factor
x^3 - 9x^2 + 15x + 25
5 - 9 = - 4
-4*5 + 15 = - 5
-5*5 + 25 = 0
so 5 is a root and (x - 5) is a factor. The other factor must be (x - 5) as well to get a product of + 25.
(x + 1)(x - 5)^2

(In practice I use my spreadsheet, since once you set up the function, it can calculate all these in the blink of an eye.)

Answer 3

Factorising cubic expressions quickly is a matter of luck.
(And of course there is no guarantee that they factorise nicely at all).
You try the factors of the final number, both positive and negative to see whether they make the expression zero.
With both of these you find that x = -1 makes them zero. This means that both have a factor of (x + 1). You divide this out from each and see what happens to the quadratic that is left. Again in both cases the quadratic factorises easily so both have three zeros which are integers. (In the second case one of the zeros is repeated so I suppose you could say that it only has two zeros.)