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 \(\geq 2\)

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TopNewesti) 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 · 3 years, 8 months ago

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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 · 3 years, 8 months ago

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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 · 3 years, 8 months ago

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