Question

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

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:
$$\begin{gathered} i^1=i \\ {i}^2=-1 \\ {i}^3=-i \\ {i}^4=1 \end{gathered}$$

Complete step-by-step answer:
The divisor is x2 + 1 so, the values of x obtained are:
$$\begin{gathered} x^2+1=0 \\ {x}^2=-1 \\ x=i,-i \end{gathered}$$
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 $5x^4+4x^3+3x^2+\text{M}\text{.x + N}$ we get for x = i
$5i^4+4i^3+3i^2+Mi+N$
$5+(-4i)+(-3)+Mi+N$
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.