well, if f(-2)=0, x+2 has to be a factor right?
oh yeah i never realized that thanks for pointing that out for me
the other factor has one degree lower than the given cuz x+2 is first order
which means the factor is (x+2)(ax^2+bx+c) and a is not 0
now set two equal to each other, expand LHS and compare coefficients to determine a,b,c if the secondary polynomial has real roots, rewrite the polynomial as (x+x1)(x+x2) where x1and x2 are the roots
im lost now
Ok lets work this out, you undertand that x+2 is one of the factors right
and the other part has the form ax^2+bx+c,right?
where did you get the ax and the bx and the c
Do you mean how did I come up with this form?
well, 5^3 can be factored into 5^2 and 5, right
same thing here, we pull a x out of x^3,the left part has to be x^2
so we have the answer already, it is (x+2)(ax^2+bx+c)
therefore it has to equal to x^3-3x^2-16x-12
so whats the answer
we are getting there, still a long way to go actually
so we have (x+2)(ax^2+bx+c)=x^3-3x^2-16x-12
now we have a=1,2c=-12,2a+b=-3, so you can solve for a b and c
so tell me what are a b and c
now set ax^2+bx+c=0, is there any real root?
well, yes actually b^2-4ac text remember?
so (-5)^2-4*-6=49, which is neat, because it is 7^2
now can you solve the equation?
so what is the solution?
I am the quadratic equation what are the roots, there are two, if you haven't realized already
well thanks for the help i got to go now