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

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

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

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

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

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## Comments

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

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That's one possible approach. However, as $d(x)$ 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?Log in to reply

$f(x) = x^n$

$f(1) = 1$

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

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

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

$f'(1) = n$

$Q(1) = n$

$Q(x) = (x - 1)q(x) + n$

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

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

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

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

$f(-1) = 1$

$4a - 19 = 1$

$a = 5$

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Yes, that is the right approach. More generally,

What does this remind you of?Hint:What is the remainder when $f(x)$ is divided by $(x-a) ^3$@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.

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$x^n = ((x - a) + a)^n$

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What is the general solution, and why?

Find the "one-line" explanation for it.

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

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$f(x)$ is divided by $(x-a)^n$?

What is the remainder whenWhat is a one-line explanation for the answer?

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$x^n - (x - a)^n$

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Wait, IM SO CLOSE I MIGHT HAVE FIGURED IT OUT

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

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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.

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Great observation about using LaGrange polynomials for distinct roots!

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