Question

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

Verified
149.7k+ views
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.