Question

If ${x^{31}} + 31$ is divided by $x + 1$, then what will be the remainder?

Answer 1

Hint: We will use the remainder theorem, in which we equate the linear factor with zero and then substituting the value in the given function. Also, the odd power of $\left( { - 1} \right)$ is equal to $ - 1$.

Complete step-by-step solution:
We know from that it is given in the question that the polynomial function is $f\left( x \right) = {x^{31}} + 31$ and the linear factor is $x + 1$.
According to the remainder theorem, when a polynomial function $f\left( x \right)$ is divided a linear factor $\left( {x - a} \right)$, then the remainder of the given function is given by $f\left( a \right)$.
Now we apply the remainder theorem in the given question, in the first equation the given linear factor equal to zero to obtain the value of $x$
$x + 1 = 0\\
x = - 1$
Now we substitute the value of $x$ as $ - 1$ in the given polynomial function, we get,
$f\left( { - 1} \right) = {\left( { - 1} \right)^{ - 31}} + 31\\
 = - 1 + 31\\
 = 30$

Hence, the value of the remainder of the given polynomial function is 30.

Additional Information: The polynomial is an expression that involves variables, power and constants and the polynomial function is the function that contains only positive integers as power of the variable in different types of equations like linear, cubic, etc. For example, $3{x^2} + 6$ is a polynomial function donations exponent 2.

Note: Remainder theorem is the easy and short method to calculate the remainder of a given polynomial function. It states that if a polynomial $f\left( x \right)$ is divided by the linear factor $\left( {x - a} \right)$, then the remainder of the function is $f\left( a \right)$ that means the remainder after performing the systematic division is same when you replace $a$ in the given function.