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
Verified
479.1k+ views
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.
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.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 English: Engaging Questions & Answers for Success
Trending doubts
Assertion The planet Neptune appears blue in colour class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The term disaster is derived from language AGreek BArabic class 10 social science CBSE
Imagine that you have the opportunity to interview class 10 english CBSE
10 examples of evaporation in daily life with explanations
Differentiate between natural and artificial ecosy class 10 biology CBSE