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Hint: Remainder theorem is used to determine the remainder when two polynomials are given in the question for the division. The remainder theorem states that when a polynomial\[g\left( x \right)\] is divided by a linear polynomial, \[\left( {x - a} \right)\] then we get the remainder as \[g(a)\]and quotient as $q(x)$. The remainder is an integer that is left behind when a dividend is divided by the divisor. This theorem is applicable only when the polynomial is divided by a linear polynomial. A linear polynomial is generally represented as \[f\left( x \right) = ax + b\]where \[b\] is the constant term, whereas polynomial is the expression which consists of variables and coefficients having the operations of addition, multiplication, subtraction, and exponent. A polynomial function in degree n can be written as \[g(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + {a_{n - 2}}{x^{n - 2}} + ........... + {a_2}{x^2} + {a_1}{x^1} + {a_0}\].
Another method to find the remainder is by dividing the dividend by the divisor, which is the general method to find the remainder, \[{\text{R = }}\dfrac{{{\text{dividend}}}}{{{\text{divisor}}}}\]
Complete step by step solution: Here \[{x^4} - {a^2}{x^2} + 3x - 6a\] is the dividend to which number is to be divided and \[x + a\] is the divisor by which the number is divided.
Now equate the divisor, which is a linear polynomial with \[0\] hence we get,
\[
x + a = 0 \\
x = - a \\
\]
So we can conclude when the polynomial \[f\left( x \right) = {x^4} - {a^2}{x^2} + 3x - 6a\] is divided by the linear polynomial, \[x = - a\]we get the remainder\[f\left( { - a} \right)\],
\[
f\left( x \right) = {x^4} - {a^2}{x^2} + 3x - 6a \\
f\left( { - a} \right) = {\left( { - a} \right)^4} - {a^2}{\left( { - a} \right)^2} + 3\left( { - a} \right) - 6a \\
= {a^4} - {a^4} - 3a - 6a \\
= - 9a \\
\]
We know it \[\left( {x + a} \right)\] is the factor of \[f\left( x \right)\]then we can write \[f\left( { - a} \right) = 0\] ; hence we get
\[
f\left( { - a} \right) = 0 \\
- 9a = 0 \\
a = 0 \\
\]
So the value of a is 0.
Option A is correct.
Note: Before finding the remainder, check whether the divisor is a linear polynomial; otherwise, the division is not possible. Another method to find the remainder is by dividing the dividend by the divisor by a long division method, which includes more calculations, and the chances of getting the error are more.
Another method to find the remainder is by dividing the dividend by the divisor, which is the general method to find the remainder, \[{\text{R = }}\dfrac{{{\text{dividend}}}}{{{\text{divisor}}}}\]
Complete step by step solution: Here \[{x^4} - {a^2}{x^2} + 3x - 6a\] is the dividend to which number is to be divided and \[x + a\] is the divisor by which the number is divided.
Now equate the divisor, which is a linear polynomial with \[0\] hence we get,
\[
x + a = 0 \\
x = - a \\
\]
So we can conclude when the polynomial \[f\left( x \right) = {x^4} - {a^2}{x^2} + 3x - 6a\] is divided by the linear polynomial, \[x = - a\]we get the remainder\[f\left( { - a} \right)\],
\[
f\left( x \right) = {x^4} - {a^2}{x^2} + 3x - 6a \\
f\left( { - a} \right) = {\left( { - a} \right)^4} - {a^2}{\left( { - a} \right)^2} + 3\left( { - a} \right) - 6a \\
= {a^4} - {a^4} - 3a - 6a \\
= - 9a \\
\]
We know it \[\left( {x + a} \right)\] is the factor of \[f\left( x \right)\]then we can write \[f\left( { - a} \right) = 0\] ; hence we get
\[
f\left( { - a} \right) = 0 \\
- 9a = 0 \\
a = 0 \\
\]
So the value of a is 0.
Option A is correct.
Note: Before finding the remainder, check whether the divisor is a linear polynomial; otherwise, the division is not possible. Another method to find the remainder is by dividing the dividend by the divisor by a long division method, which includes more calculations, and the chances of getting the error are more.
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