# Thread: p(x) is divisible by (ax+b)^n

1. ## p(x) is divisible by (ax+b)^n

Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
Now my p(x) = (ax+b)^n-1 s(x)
then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?

2. Originally Posted by dan34
Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
Now my p(x) = (ax+b)^n-1 s(x)
then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?
$(3x + 2)^2 \ | \ p(x) = (3x + 2)^2 \text { and } p'(x) = 6(3x + 2).$

But it is not generally true that $(3x + 2)^{(100-1)} \ | \ 6(3x + 2).$

Tough to prove something that is false!

3. Originally Posted by dan34
Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
Now my p(x) = (ax+b)^n-1 s(x)
then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?
Your subject line refers to "(ax + b)n", rather than to "(ax + b)2". Is the square in the "show" statement a typo for a power of "n"?

Thank you!