Question

If the HCF of f(X) =(x-3) (x²+X+m)and g(X) = (x-2) (x²+X+n) is (x-3) (x-2),then m+n = ---------​

Answer 1

Answer:

If x+n is the hcf of x^2 +ax+b and X^2+cx+D then what is the value of n?

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Ved Prakash Sharma

Answered 3 years ago

x+n is HCF of x^2+ax+b and x^2+cx+d. Therefore x+n is a factor of each expression.

Put x=-n in the given expressions.

(-n)^2+a×(-n)+b=0

n^2-a.n+b=0………………….(1)

(-n)^2+c.(-n)+d=0

n^2-c.n+d=0……………………(2)

Subtract eq.(2) from (1)

c.n-a.n-d+b=0

n(c-a)=d-b

n=(d-b)/(c-a) , answer

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Kamal Joshi

Answered 4 years ago

If x + n is the hcf, then it is a common factor of both expressions. In that case we can use factor theorem which says x-a is a factor of a polynomial p(x) if p(a) = 0. So here,

(-n)^2 + a(-n)+b=0 and (-n)^2+c(-n)+d=0. Solving for n we get n = (b-d)/(c-a)

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Chhavi Rathi

Answered 4 years ago

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Dave Clark

Updated 3 years agoErmal FeleqiRavi Ranjan Kumar Singh

How would I approach ∫dxx(x+1)(x+2)(x+3)⋯(x+n)?

Integrals of rational polynomials often yield to partial fraction decomposition. In this case, we can find a useful pattern based on Pascal’s Triangle.

The key observation is:

1x(x+1)(x+2)(x+3)⋯(x+n)=10!n!x−11!(n−1)!(x+1)+12!(n−2)!(x+2)−13!(n−3)!(x+3)+⋯+(−1)nn!0!(x+n)=∑i=0n(−1)ii!(n−i)!(x+i)

∴∫1x(x+1)(x+2)(x+3)⋯(x+n)dx=∑i=0n(−1)ilog|x+i|i!(n−i)!+C

Proof of observation

We perform

Step-by-step explanation:

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