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

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

Note by Calvin Lin
2 years, 8 months ago

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

- 2 years, 8 months 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?

Staff - 2 years, 8 months 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 $$5(x - 1)^{2} + 10x - 9 = 5x^2 - 4.$$

- 2 years, 7 months 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.

Staff - 2 years, 7 months ago

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

- 2 years, 7 months ago

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".

Staff - 2 years, 7 months ago

Wait, IM SO CLOSE I MIGHT HAVE FIGURED IT OUT

- 2 years, 7 months ago

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

Staff - 2 years, 7 months ago

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

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

- 2 years, 7 months ago

What is the general solution, and why?

Find the "one-line" explanation for it.

Staff - 2 years, 7 months 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 $$\sum_{i=n-k+1}^n nC_i (x - a)^{n - i} a^i.$$

- 2 years, 7 months ago

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

What is a one-line explanation for the answer?

Staff - 2 years, 7 months ago

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

- 2 years, 7 months ago