nP3 = 60
????
so
n!/(n-3)!
=
n(n-1)(n-2)(n-3)....1
over
(n-3)(n-4)(n-5)....1
=
n(n-1)(n-2)
=
n^3-3n^2+2n-60=0
So what do you do polynomial division?
what if its like nP20
is there something Im doing wrong?
thanks.
nP3 = 60
????
so
n!/(n-3)!
=
n(n-1)(n-2)(n-3)....1
over
(n-3)(n-4)(n-5)....1
=
n(n-1)(n-2)
=
n^3-3n^2+2n-60=0
So what do you do polynomial division?
what if its like nP20
is there something Im doing wrong?
thanks.
Yep you do polynomial division, I'm not sure how you were taught. I was taught to try and find a factor.
f(5) = 125-75+10-60 = 0.
Therefore (n-5) is a factor and we can divide through using long division and solve the remaining quadratic.
or compare coefficients: (n-5)(an^2+bn+c) = n^3-3n^2+2n-60
For example compare coefficients of x^0 (the constant): -5c = -60 therefore c=12
Ive done a couple using long division, and the remainder = 0 so that is good....But the quadratics I get are not factorable nor can you use the quadratic equation because the discriminant is negative?
for the example I gave I got n^2+2n+12?
Edit.
I guess it means the answer is the one factor of 5.