Question

There is only one way to choose real number M and N such that when the polynomial $5{x^4} + 4{x^3} + 3{x^2} + {\text{M}}{\text{.x + N}}$ is divided by polynomial ${x^2} + 1$ the remainder is 0. If M and N assume these unique values then M-N:
$ {\text{A}}{\text{. 6}} \\
  {\text{B}}{\text{. - 2}} \\
  {\text{C}}{\text{. 6}} \\
  {\text{D}}{\text{. 2}} \\ $

Answer 1

Hint: Substitute i and –i in the polynomial. From these substitutions we obtain 2 expressions. By equating those expressions to 0 and making the coefficients null get the values of M and N.
Formula Used:
$
  {i^1} = i \\
  {i^2} = - 1 \\
  {i^3} = - i \\
  {i^4} = 1 \\ $

Complete step-by-step answer:
The divisor is x2 + 1 so, the values of x obtained are:
$ {x^2} + 1 = 0 \\
  {x^2} = - 1 \\
  x = i, - i $
The values of the divisor can be substituted in the polynomial as
$p(x) = q(x).g(x) + r(x)$
Where, p(x)= dividend, q(x)=quotient, g(x)=divisor and r(x)=remainder
To get the values of x we equate g(x) with 0. Now, if we substitute the value of x at LHS and RHS then, g(x) becomes 0. Hence $g(x).q(x)=0$ So, then we are left with, $p(x)=r(x)$.
Substitute these values in the polynomial $5{x^4} + 4{x^3} + 3{x^2} + {\text{M}}{\text{.x + N}}$ we get for x = i
$5{i^4} + 4{i^3} + 3{i^2} + Mi + N$
$5 + (-4i) + (-3) + Mi + N$
Collecting the terms with iota and without iota $ (M-4)i + (N+2)$
Similarly on putting $x = -i$, we get
$5{\left( { - i} \right)^4} + 4{\left( { - i} \right)^3} + 3{\left( { - i} \right)^2} + M\left( { - i} \right) + N$
$5 +4i + (-3)- Mi + N$
Collecting the terms with iota and without iota- $(-M+4)i + (N+2)$
Which is $[- (M-4)i] + [(N+2)]$
So, we observe that the 2 expressions are just the same. So, just equating to 0 as is given in the question that the remainder must be 0.
$(M-4)i + (N+2)=0$
We have $(M-4)i = 0$, gives $M=4$ and $(N+2)=0$ gives $N= -2$.
$M=4$ and $N= -2$ are the required values.

Note: Another way to solve the same numerical is by dividing the given polynomial by ${x^2} + 1$ and then equating the remainder obtained to 0 as is stated in the question.
But division a polynomial by polynomial will be a longer approach that’s why we don’t consider this method.