Answer
Verified
459.9k+ 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
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE