Remainder Theorem - Follow on

f(x)=q(x)d(x)+r(x) f(x) = q(x) d(x) + r(x)

When d(x)d(x) is a polynomial with no repeated roots, Prasun has stated how we could determine r(x)r(x) using the Remainder-Factor Theorem.

How should we deal with the case when d(x)d(x) has repeated roots?


What is the remainder when x10 x^{10} is divided by (x1)2 (x-1) ^2 ?

What is the remainder when x10 x^{10} is divided by (x1)2(x+1) (x-1) ^2 (x+1) ?

Note by Calvin Lin
4 years ago

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Differentiate throughout and evaluate the expression at the value of the repeated root.

Sudeep Salgia - 4 years ago

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That's one possible approach. However, as d(x)d(x) gets complicated, you could run into various issues. E.g. how would you evaluate the 2 given questions?

Is there another simpler way?
Hint: How can we apply partial fractions cover up rule to repeated roots?

Calvin Lin Staff - 4 years ago

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f(x)=xnf(x) = x^n

f(1)=1f(1) = 1

f(x)=(x1)Q(x)+1f(x) = (x - 1)Q(x) + 1

f(x)=Q(x)+(x1)Q(x)f'(x) = Q(x) + (x - 1)Q'(x)

f(x)=nxn1f'(x) = nx^{n-1}

f(1)=nf'(1) = n

Q(1)=nQ(1) = n

Q(x)=(x1)q(x)+nQ(x) = (x - 1)q(x) + n

f(x)=(x1)((x1)q(x)+n))+1f(x) = (x - 1)((x - 1)q(x) + n)) + 1

f(x)=(x1)2q(x)+nxn+1f(x) = (x - 1)^{2}q(x) + nx - n + 1

The remainder when x10 x^{10} is divided by (x1)2(x - 1)^{2} is 10x9.10x - 9.

f(x)=(x1)2(x+1)q1(x)+a(x1)2+nxn+1f(x) = (x - 1)^{2}(x + 1)q_1(x) + a(x - 1)^{2} + nx - n + 1

f(x)=x10f(x) = x^{10}

f(1)=1f(-1) = 1

4a19=14a - 19 = 1

a=5a = 5

The remainder when x10 x^{10} is divided by (x1)2(x+1)(x - 1)^{2}(x + 1) is 5(x1)2+10x9=5x24.5(x - 1)^{2} + 10x - 9 = 5x^2 - 4.

汶良 林 - 4 years ago

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Yes, that is the right approach. More generally,

The remainder when f(x) f(x) is divided by (xa)2 (x-a) ^2 is f(a)(xa)+f(a) f'(a) (x-a) + f(a) .

What does this remind you of?

Hint: What is the remainder when f(x) f(x) is divided by (xa)3 (x-a) ^3

@Sudeep Salgia @Prasun Biswas The above is a useful way to remember how to find the remainder when divided by a linear power. It forms the linkage between remainders of a polynomial, and how the polynomial looks like at a point.

Calvin Lin Staff - 4 years ago

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Tangent line to f(x)=xnf(x) = x^n at (a,f(a)(a, f(a)?

汶良 林 - 4 years ago

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@汶良 林 Close, but I'm thinking of something deeper. What would the answer to the hint be? That should be a 1 line answer with the correct "reminder".

Calvin Lin Staff - 4 years ago

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xn=((xa)+a)nx^n = ((x - a) + a)^n

The remainder when x10 x^{10} is divided by (xa)3(x - a)^3 is nC2(xa)2an2+nC1(xa)1an1+an._nC_2(x - a)^2 a^{n - 2} + _nC_1(x - a)^1 a^{n - 1} + a^n .

汶良 林 - 4 years ago

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What is the general solution, and why?

Find the "one-line" explanation for it.

Calvin Lin Staff - 4 years ago

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((xa)+a)n=i=0nnCi(xa)niai(( x - a) + a)^n = \sum_{i=0}^n nC_i (x - a)^{n - i} a^i

The remainder when xn x^n is divided by (xa)k(x - a)^k is i=nk+1nnCi(xa)niai. \sum_{i=n-k+1}^n nC_i (x - a)^{n - i} a^i.

汶良 林 - 4 years ago

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@汶良 林 What is the remainder when f(x) f(x) is divided by (xa)n (x-a)^n ?

What is a one-line explanation for the answer?

Calvin Lin Staff - 4 years ago

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@Calvin Lin xn(xa)nx^n - (x - a)^n

汶良 林 - 4 years ago

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Wait, IM SO CLOSE I MIGHT HAVE FIGURED IT OUT

CS ಠ_ಠ Lee - 4 years ago

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Please avoid typing in all capital letters, as that is considered rude on the internet.

Calvin Lin Staff - 4 years ago

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Here is a neat trick that does not involve "differentiation" or tangents at all. All you need is to know how Lagrange polynomials work.

EDIT: This method does not take into account roots of multiple multiplicity. I am not sure how the formula will work in such a scenario. Comments welcome.

A Brilliant Member - 10 months ago

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Great observation about using LaGrange polynomials for distinct roots!

Calvin Lin Staff - 10 months ago

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