Hello everyone!

We all know that by the Remainder theorem, If \(p(x)\) is divided by \((x-a)\), The remainder is \(p(a)\)

But, What is the remainder when a polynomial is divided by a quardatic polynomial?

Can someone Help me regarding this? Thanks

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TopNewestSince you raised this general question and I wanted to get a general procedure for this, I spent almost all of my evening thinking about this. This is what I came up with. Maybe this might help you.

Consider the polynomial \(p(x)\) of degree \(d_p\) and the divisor polynomial \(d(x)\) of degree \(d_q\) such that \(d_q\leq d_p\) (obviously) where we have \(d_p,d_q\in\Bbb{Z_0^+}\). Denote by \(q(x)\) the quotient polynomial you get while performing division algorithm and denote by \(r(x)\) the remainder you get when \(p(x)\) is divided by \(d(x)\).

Note:If \(d_p=0\), then the polynomial is a constant function which is divisible by any divisor polynomial of degree \(0\). Also, if we have \(d_q=0\), we simply have that the divisor is a constant function which obviously divides polynomials \(p(x)\) of any degree. If \(d_p=d_q=0\), the result is the same which follows from the already discussed two cases. So, these trivial cases can be dismissed from our examination since they always give the remainder as \(0\).We proceed to get a systematic procedure to deal with the non-trivial cases, i.e., when \(d_p,d_q\in\Bbb{Z^+}\).

Consider the set \(\large\{r_i\}_{i=1}^{i=q_d}\) of all roots of \(d(x)\). By division algorithm, we have,

\[p(x)=q(x)d(x)+r(x)\implies p(x)=\left(q(x)\cdot\large\prod_{k=1}^{q_d}(x-r_k)\right)+r(x)\]

Obviously, \(r(x)\) is a polynomial of degree \(q_d-1\) and hence is of the form \(\displaystyle\sum_{k=0}^{q_d-1}a_kx^k\) where the sequence \(\{a_k\}_{k=0}^{k=q_d-1}\) is a sequence of constants depending upon \(d(x)\). We then obtain,

\[\large p(r_k)=r(r_k)=\sum_{j=0}^{q_d-1}a_j(r_k)^j~\forall~k\in\Bbb{Z^+_{\leq q_d}}\]

So, from the above result, you get \(q_d\) equations for a system with \(q_d\) unknowns. Hence, you can bash it out using methods of solving large system of equations, mostly methods from Linear Algebra (matrices, etc).

There's yet another way to finish it off elegantly

if the roots of \(d(x)\) form an arithmetic progression. The above result also can be said to give us \(q_d\) different polynomial values for \(r(x)\) which has degree \(q_d-1\). You can use method of differences now to get the final difference column value easily and then use the "reconstruction formula" in method of differences to get the polynomial \(r(x)\), which is your answer. And, we are done. \(_{\square}\)Note that the form we took for \(d(x)\) accounts for all cases since any other kind of structure of \(d(x)\) can easily be obtained by multiplying a scalar to \(q(x)\) in the division algorithm which doesn't affect \(r(x)\).

Since I already wrote so long for this, let's take the time to state an explicit example.

Example:Find the remainder when \(p(x)=x^{10}\) is divided by the cubic polynomial \(d(x)=(x-1)(x-2)(x-3)\).Solution:We have, through our procedure as shown above that,\[p(1)=r(1)=1^{10}=1\quad\textrm{and}\quad p(2)=r(2)=2^{10}=1024\quad\textrm{and}\quad p(3)=r(3)=3^{10}=59049\]

Now, construct the difference table for \(r(x)\) as follows:

\[\begin{array}{|c|c|c|c|c|}\hline x&r(x)&D_1(x)&D_2(x)\\ \hline 1&1&1023&57002\\ 2&1024&58025 \\ 3&59049\\ \hline\end{array}\]

Now, we do the reconstruction using the following formula:

\[r(x)=r(1)+\sum_{i=1}^2\left(\frac{D_i(1)}{i!}\cdot \prod_{j=1}^i(x-j)\right)=28501x^2-84480x+55980\]

As you can infer, this becomes more and more tedious as the degrees of \(p(x)\) and \(d(x)\) increase, even using method of differences or other interpolation methods.

Note:This method works only for those \(d(x)\) which has no repeated roots. As of now, this is not much of a generalization since it doesn't account for cases when \(d(x)\) has a repeated root.Log in to reply

@Calvin Lin, would you mind verifying this explanation? I don't get why would someone downvote it? Is there any flaw in the explanation? If so, would you kindly point it out?

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I don't know how one can downvote such magnificent work of generalization.I have upvoted it :)

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Firstly, the following statement is false

It is true only if the roots form an arithmetic progression.

Secondly, you have only dealt with the case where the dividend has no repeated roots, so it's not quite the "generalized polynomial".

Otherwise, it looks mostly good.

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. Let me try : Let the divisor be \(ax^2+bx+c\) whose roots are \(u,v\). So its as if , \(P(x)\) is divided by \((x-u)(x-v)\).Since the divisor has a degree of \(2\) , the remainder must have degree of 1 and thus it has the form : \(kx+m\).By division algorithm we can write :

\[P(x)=Q(x) . (x-u)(x-v) + kx+m\]

When \(P(x)\) is divided by \(x-u\) , the remainder is \(P(u)\) and When \(P(x)\) is divided by \(x-v\) , the remainder is \(P(v)\). So , to find the required remainder we must solve the following system of equations:

\[ P(u)=ku+m \\ P(v)=kv+m\]

Cheers!

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Thanks so much!

But, If we have to find a single remainder, What should we do?

For example, What will be the remainder if \({x}^{2015}\) is divided by \({x}^{2}-4x+3\) ?

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Solve system:

\[3^{2015} = 3a+b \\ 1 = a+b\]

And the remainder will be \(ax+b\)

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\[x^{2015}=(x-1)(x-3)\times q(x)+mx+b\\ \Longrightarrow 1^{2015}=1=m+b\\ 3^{2015}=3m+b\\ \Longrightarrow 2m=3^{2015}-1(by\ subtracting '3m+b'\ 'm+b')\\ \Longrightarrow m=\dfrac{3^{2015}-1}{2}\\ \Longrightarrow b=1-\dfrac{3^{2015}-1}{2}\\ \Longrightarrow remainder\ when\ x^{2015}\ is\ divided\ by\ x^2-4x+3\ is\\ \dfrac{3^{2015}-1}{2}\times x+1-\dfrac{3^{2015}-1}{2}\]. Cheers!

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Don't ask our questions on B'lliant. This is wrong. We already have mentioned what to do. So Please.

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P.S. Kindly improve your language a tad bit ;)

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And sorry .

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The divisor is a quadratic polynomial,

nota quadratic equation. The divisor cannot be an equation. The`=0`

part is invalid. You should remove that. The rest of your comment seems okay to me.EDIT: @Nihar Mahajan, I just noticed that if you're going to keep the divisor as \(ax^2+bx+c\), then the format \((x-u)(x-v)\) doesn't suite well since that forces \(a=1\). This isn't that much of a problem actually since it is corrected simply by multiplication of quotient polynomial by a constant, but for the sake of clarity, you might want to reflect this in your comment and edit it accordingly.

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Sorry , it has become an habit to write \(ax^2+bx+c=0\). :P

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@Nihar Mahajan @Manish Dash @Rajdeep Dhingra @Prasun Biswas

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How did this question come to your mind! It was my doubt! Thanks for raising this question, mehul!

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Welcome bro.

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For an explicit example, see Remainder Factor Theorem - Intermediate.

After you understood what to do, could you add in a paragraph of explanation?

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Sure sir! I will do so :)

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