# Find the conditions that ${x^3} + p{x^2} + qx + r$ may be divisible by ${x^2} + ax + b$.

Last updated date: 16th Mar 2023

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Answer

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Hint: When we divide a third degree polynomial by a second degree polynomial we will get the quotient in first degree.

Let $f(x) = {x^3} + p{x^2} + qx + r$

Here, $f\left( x \right)$ is our dividend and we have to find the conditions for ${x^2} + ax + b$ to be the divisor.

When ${x^3} + p{x^2} + qx + r$ is divisible by ${x^2} + ax + b$ , Then the result will be in unit power of x.

Let result is x+n

$ \Rightarrow \frac{{{x^3} + p{x^2} + qx + r}}{{{x^2} + ax + b}} = x + n$

So

$ \Rightarrow x({x^2} + ax + b) + n({x^2} + ax + b) = {x^3} + p{x^2} + qx + r$

$ \Rightarrow {x^3} + a{x^2} + bx + n{x^2} + anx + bn = {x^3} + p{x^2} + qx + r$

Comparing the x powers on both sides, we get

$\eqalign{

& (a + n) = p...(i) \cr

& (b + an) = q...(ii) \cr

& bn = r...(iii) \cr} $

From equations (i) and (iii) we get $a,b$ values

$\eqalign{

& b = \frac{r}{n} \cr

& a = p - n \cr} $

Substituting $a,b$ in equation (ii)

$\frac{r}{n} + n(p - n) = q$

$ \Rightarrow np - {n^2} + \frac{r}{n} = q$, where n is real number

This $np - {n^2} + \frac{r}{n} = q$ is the required condition that ${x^3} + p{x^2} + qx + r$ is divisible by ${x^2} + ax + b$.

Note: Division of polynomials may look difficult, but it is very similar to the division of real numbers. Polynomial long division is a method for dividing a polynomial by another polynomial of a lower degree. If $P\left( x \right)$ and $D\left( x \right)$ are polynomials, with $D\left( x \right) \ne 0,$then there exist unique polynomial $Q(x)$ and $R(x),$ where $R(x)$ is either 0 or of degree less than the degree of $D(x)$ , such that

$\frac{{P(x)}}{{D(x)}} = Q(x) + \frac{{R(x)}}{{D(x)}}$ Or $P(x) = D(x).Q(x) + R(x)$

The polynomials $P\left( x \right)$ and $D\left( x \right)$ are called dividend and divisor respectively, $Q(x)$ is the quotient, and $R(x)$ is the remainder.

Let $f(x) = {x^3} + p{x^2} + qx + r$

Here, $f\left( x \right)$ is our dividend and we have to find the conditions for ${x^2} + ax + b$ to be the divisor.

When ${x^3} + p{x^2} + qx + r$ is divisible by ${x^2} + ax + b$ , Then the result will be in unit power of x.

Let result is x+n

$ \Rightarrow \frac{{{x^3} + p{x^2} + qx + r}}{{{x^2} + ax + b}} = x + n$

So

$ \Rightarrow x({x^2} + ax + b) + n({x^2} + ax + b) = {x^3} + p{x^2} + qx + r$

$ \Rightarrow {x^3} + a{x^2} + bx + n{x^2} + anx + bn = {x^3} + p{x^2} + qx + r$

Comparing the x powers on both sides, we get

$\eqalign{

& (a + n) = p...(i) \cr

& (b + an) = q...(ii) \cr

& bn = r...(iii) \cr} $

From equations (i) and (iii) we get $a,b$ values

$\eqalign{

& b = \frac{r}{n} \cr

& a = p - n \cr} $

Substituting $a,b$ in equation (ii)

$\frac{r}{n} + n(p - n) = q$

$ \Rightarrow np - {n^2} + \frac{r}{n} = q$, where n is real number

This $np - {n^2} + \frac{r}{n} = q$ is the required condition that ${x^3} + p{x^2} + qx + r$ is divisible by ${x^2} + ax + b$.

Note: Division of polynomials may look difficult, but it is very similar to the division of real numbers. Polynomial long division is a method for dividing a polynomial by another polynomial of a lower degree. If $P\left( x \right)$ and $D\left( x \right)$ are polynomials, with $D\left( x \right) \ne 0,$then there exist unique polynomial $Q(x)$ and $R(x),$ where $R(x)$ is either 0 or of degree less than the degree of $D(x)$ , such that

$\frac{{P(x)}}{{D(x)}} = Q(x) + \frac{{R(x)}}{{D(x)}}$ Or $P(x) = D(x).Q(x) + R(x)$

The polynomials $P\left( x \right)$ and $D\left( x \right)$ are called dividend and divisor respectively, $Q(x)$ is the quotient, and $R(x)$ is the remainder.

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