Question

Divide the given polynomial: $7{x^4} + 4{x^2} + 3x - 5$ by $2{x^2} + 3x - 2$.

Answer 1

In this particular question use the concept that if we divide fourth-order polynomial by second-order polynomial the quotient will be a second-order polynomial so assume any general second-order polynomial be the quotient (say, $p{x^2} + q x + r$) where p, q, and r are the constant real parameters, and there will be a remainder also say (ax + b), where a and b are constant real parameters so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given polynomial
$f\left( x \right) = 7{x^4} + 4{x^2} + 3x - 5$
Now when this polynomial is divided by another polynomial, $2{x^2} + 3x - 2$, the quotient will always be a second order polynomial.
Let, the quotient be $p{x^2} + q x + r$, where p, q, and r are constant real parameters and the remainder be (ax + b), where a and b are constant real parameters
$\Rightarrow \dfrac{{f\left( x \right)}}{D} = Q\dfrac{R}{D}$
Where, Q = quotient, R = remainder, and D = divisor.
Now substitute the value of f (x) we have,
$\Rightarrow \dfrac{{7{x^4} + 4{x^2} + 3x - 5}}{{2{x^2} + 3x - 2}} = \left( {p{x^2} + qx + r} \right)\dfrac{{\left( {ax + b} \right)}}{{2{x^2} + 3x - 2}}$
$\Rightarrow 7{x^4} + 4{x^2} + 3x - 5 = \left( {p{x^2} + qx + r} \right)\left( {2{x^2} + 3x - 2} \right) + \left( {ax + b} \right)$
Now simplify it we have,
$\Rightarrow 7{x^4} + 4{x^2} + 3x - 5 = \left( {2p{x^4} + \left( {3p + 2q} \right){x^3} + \left( { - 2p + 3q + 2r} \right){x^2} + \left( { - 2q + 3r + a} \right)x - \left( {2r - b} \right)} \right)$
Now on comparing we have,
$\Rightarrow 2p = 7$..................... (1)
$\Rightarrow 3p + 2q = 0$................ (2)
$\Rightarrow - 2p + 3q + 2r = 4$................ (3)
$\Rightarrow - 2q + 3r + a = 3$............. (4)
$\Rightarrow 2r - b = 5$............... (5)
So from equation (1) we have,
$\Rightarrow p = \dfrac{7}{2}$
Substitute this value in equation (2) we have,
$\Rightarrow 3\left( {\dfrac{7}{2}} \right) + 2q = 0$
$\Rightarrow 2q = - 3\left( {\dfrac{7}{2}} \right)$
$\Rightarrow q = \left( { - \dfrac{{21}}{4}} \right)$
From equation (3) we have,
$\Rightarrow - 2p + 3q + 2r = 4$
$\Rightarrow - 2\left( {\dfrac{7}{2}} \right) + 3\left( {\dfrac{{ - 21}}{4}} \right) + 2r = 4$
$\Rightarrow 2r = 4 + 7 + \left( {\dfrac{{63}}{4}} \right) = \dfrac{{107}}{4}$
$\Rightarrow r = \dfrac{{107}}{8}$
From equation (4) we have,
$\Rightarrow - 2q + 3r + a = 3$
$\Rightarrow - 2\left( {\dfrac{{ - 21}}{4}} \right) + 3\left( {\dfrac{{107}}{8}} \right) + a = 3$
$\Rightarrow a = 3 - \left( {\dfrac{{21}}{2}} \right) - \left( {\dfrac{{321}}{8}} \right) = \dfrac{{ - 381}}{8}$
From equation (5) we have,
$\Rightarrow 2r - b = 5$
$\Rightarrow 2\left( {\dfrac{{107}}{8}} \right) - b = 5$
$\Rightarrow \left( {\dfrac{{107}}{4}} \right) - 5 = b$
$\Rightarrow b = \dfrac{{87}}{4}$
So the quotient polynomial is
$\Rightarrow p{x^2} + qx + r = \dfrac{7}{2}{x^2} - \dfrac{{21}}{4}x + \dfrac{{107}}{8}$
And the remainder polynomial is
$\Rightarrow ax + b = \dfrac{{ - 381}}{8}x + \dfrac{{87}}{4}$
$\Rightarrow \dfrac{{7{x^4} + 4{x^2} + 3x - 5}}{{2{x^2} + 3x - 2}} = \left( {\dfrac{7}{2}{x^2} - \dfrac{{21}}{4}x + \dfrac{{107}}{8}} \right)\dfrac{{\left( {\dfrac{{ - 381}}{8}x + \dfrac{{87}}{4}} \right)}}{{2{x^2} + 3x - 2}}$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that when we multiply the divisor polynomial by the quotient polynomial and in this add the remainder polynomial we will get the resultant polynomial, so on comparing we get some equation with unknown parameters, simply solve these equations we will get the required answer.