# Division of Polynomial by quadratic

There was a question in which it was asked to find the remainder when $$x^{100}$$ is divided by $$x^2-3x+2$$. However I somehow manged to solve this question. But then I thought what would be the general case of this; that is---

What is the remainder when $$f(x)$$ is divided by

i) $$(x-a)(x-b)$$

ii) $$ax^2+bx+c$$

(Provided that $$f(x)$$ is of degree $$\geq 2$$

Note by Kishlaya Jaiswal
4 years, 9 months ago

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i) Let $$Q(x)$$ and $$R(x)$$ be the quotient and remainder respectively when $$f(x)$$ is divided by $$(x-a)(x-b)$$.

Write $$f(x) = (x-a)(x-b)Q(x)+R(x)$$. Then, $$f(a) = R(a)$$ and $$f(b) = R(b)$$.

Since $$R(x)$$ must be a linear polynomial, we have:

$$R(x) = \dfrac{b-x}{b-a}f(a) + \dfrac{x-a}{b-a}f(b) = \dfrac{f(b)-f(a)}{b-a}x + \dfrac{bf(a) - af(b)}{b-a}$$.

Both expressions are equivalent, pick the one you like best.

ii) Write $$ax^2+bx+c = a(x-\alpha)(x-\beta)$$, then use part i).

- 4 years, 9 months ago

Sorry, I didn't understand how did you got the expression for $$R(x)$$

But this is how I solved it -

$f(x) = q_1(x-a) + r_1 \Rightarrow \frac{f(x)}{(x-a)} = q_1 + \frac{r_1}{(x-a)} ...(i)$

$f(x) = q_2(x-b) + r_2 \Rightarrow \frac{f(x)}{(x-b)} = q_2 + \frac{r_2}{(x-b)} ...(ii)$

where $$r_1 = f(a) , r_2 = f(b)$$

Now subtracting the equation (i) from equation (ii) we get -

$f(x)\frac{(b-a)}{(x-a)(x-b)} = (q_1 - q_2) + \frac{(r_2x - r_1x + r_1b - r_2a)}{(x-a)(x-b)}$

Therefore the remainder when $$f(x)$$ is divided by $$(x-a)(x-b)$$ must be $\frac{(r_2x - r_1x + r_1b - r_2a)}{b-a}$

Placing the values of $$r_2 = f(b)$$ and $$r_1 = f(a)$$, we get -

$\frac{f(b) - f(a)}{b-a}x + \frac{bf(a) - af(b)}{b-a}$

I didn't understand that why are we getting the quotient as $$q_1 - q_2$$

(Please tell me if I went wrong somewhere)

- 4 years, 9 months ago

Calculation of Remainder when $$x^{100}$$ is divided by $$x^2-3x+2$$

We know that when $$p(x)$$ is divided by $$g(x)$$. Then $$q(x)$$ is quotient and $$r(x)$$ is remainder

Where Degree of $$r(x)$$ is less then $$q(x)$$

So Using Division algorithm, $$p(x) = q(x)\cdot g(x)+r(x)$$

So here $$p(x) = x^{100}$$ and $$g(x) = (x^2-3x+2) = (x-1)\cdot (x-2)$$ and Let $$r(x) = ax+b$$

So Degree of $$r(x) <$$ Degree of $$g(x)$$

So $$x^{100} = (x-1) \cdot (x-2)\cdot g(x)+ax+b.................................................[1]$$

Put $$x-1 = 0\Rightarrow x = 1$$

$$1 = 0+a+b............................................[2]$$

Put $$x -2 = 0\Rightarrow x = 2$$

$$2^{100} = 0+2a+b...................................[3]$$

Now solving $$[2]$$ and $$[3]$$, we get

$$a = 2^{100} - 1$$ and $$b = 2-2^{100}$$

So $$r(x) = (2^{100}-1)\cdot x +(2-2^{100})$$

- 4 years, 9 months ago