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 $\geq 2$

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

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).

Jimmy Kariznov - 7 years, 8 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)

Kishlaya Jaiswal - 7 years, 8 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 $r(x) = (2^{100}-1)\cdot x +(2-2^{100})$

jagdish singh - 7 years, 8 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}`	$\boxed{123}$