Remainder Theorem - Follow on

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

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 $(x-1) ^2 (x+1)$ ?

#Algebra

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Comments

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

Sudeep Salgia - 6 years ago

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?

Calvin Lin Staff - 6 years ago

$f(x) = x^n$

$f(1) = 1$

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

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

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

$f'(1) = n$

$Q(1) = n$

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

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

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

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

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

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

$f(-1) = 1$

$4a - 19 = 1$

$a = 5$

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

汶良林 - 6 years ago

Yes, that is the right approach. More generally,

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

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.

Calvin Lin Staff - 6 years ago

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

汶良林 - 6 years ago

@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 - 6 years ago

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

The remainder when $x^{10}$ is divided by $(x - a)^3$ is $_nC_2(x - a)^2 a^{n - 2} + _nC_1(x - a)^1 a^{n - 1} + a^n .$

汶良林 - 6 years ago

What is the general solution, and why?

Find the "one-line" explanation for it.

Calvin Lin Staff - 6 years ago

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

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

汶良林 - 6 years ago

@汶良林 – What is the remainder when $f(x)$ is divided by $(x-a)^n$ ?

What is a one-line explanation for the answer?

Calvin Lin Staff - 6 years ago

@Calvin Lin – $x^n - (x - a)^n$

汶良林 - 6 years ago

Wait, IM SO CLOSE I MIGHT HAVE FIGURED IT OUT

CS ಠ_ಠ Lee - 6 years ago

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

Calvin Lin Staff - 6 years ago

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 Former Brilliant Member - 2 years, 9 months ago

Great observation about using LaGrange polynomials for distinct roots!

Calvin Lin Staff - 2 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$