Remainder Theorem? (Algebra 2)?
I should have added this question to the previous one, but this one just came up and I have no idea what this is...
On my textbook it describes the theorem as: 'Let P(x) be a polynomial of positive degree n. Then for any number c, P(x)= Q(x) X (x-c) + P(c) where Q(x) is a polynomial of degree n-1
I'm really confused about the sudden appearance of this 'degree' What does this have to do with anything? And how would you explain Q(x)'s definition of n-1 ?
Comments
P(x) is the equation you are dividing
Q(x) is the quotient(i think) and (x-c) is the factor
P(c) remainder
equation = (quotient * factor) + remainder
equation/factor = quotient + remainder
if divisible remainder = 0
so equation/factor = quotient -----> (x^2-1) / (x-1) = (x+1)
Any one else got better description?
once you're looking in basic terms for remainders for the 1st 3, why could you like synthetic branch? You did not ask for the quotient. in case you like the remainders in basic terms, merely plug in 3, -a million, and a million into the respective polynomials. For the 2nd area, you're searching for a polynomial with degree 3 such that f(2) = 3. using the least perplexing one, y = x^3 - 5.
The equation x² + kx + c has positive degree 2.
The equation x^n + kx + c has positive degree n.
Those formal definitions should be banned!
1 wish more people would ask me that:
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P(x)= Q(x) * (x-c) + P(c)
si divides
P(x) |_(x-c)__
r(x). . .Q(x)
donde: grad(P(x))=n
grad(r(x))<grad(x-c)=1
==>grad(r(x))=0
r(x) es una constante
==> P(x)=(x-c)*Q(x)+r(x)
P(c)=(c-c)*Q(c)+r(c)
P(c)=r(c)
==>P(x)=(x-c)*Q(x)+P(c)
puesto que P(c) es una constante