Question & Answer

Using the remainder theorem, find the remainder when \[4{{x}^{3}}+5x-10\] is divisible by \[\left( x-3 \right)\].
A. 12
B. 18
C. 96
D. 113

ANSWER Verified Verified
Hint: Use the remainder theorem. We have been given f(x) and d(x). Thus using a long division method for polynomials, find the quotient and remainder of the given f(x).

Complete step by step answer:
The remainder theorem states that when a polynomial f(x) is divided by a linear polynomial \[\left( x-a \right)\] then the remainder of that division will be equal to f(a).
If you want to evaluate the function f(x) for a given number a, you can divide the function by \[\left( x-a \right)\] and your remainder will be equal to f(a).
Now we have been given \[f(x)=4{{x}^{3}}+5x-10\] to be divided by \[\left( x-3 \right)\] by long division method.
For e.g.:- if we are dividing 7 by 2, we get remainder 1.
\[7\div 2=3\], R = 1.
7 divided by 2 equals 3 with remainder 1, where 7 is dividend, 2 is divisor, 3 is quotient and 1 is remainder.
\[\therefore 7=2\times 3+1\].
\[\therefore f(x)\div d(x)=q(x)\] with a remainder r(x).
Thus the polynomials can be written in this form.
\[\therefore f(x)=4{{x}^{3}}+5x-10\] and \[d(x)=x-3\]

First find \[4{{x}^{2}}(x-3)=4{{x}^{3}}-12{{x}^{2}}\] as to get the first term. Then we get the remainder as \[\left( 12{{x}^{2}}+5x \right)\]. Now multiply \[12x\left( x-3 \right)\] to get \[12{{x}^{2}}-36x.\] We get the remainder as \[\left( 41x-10 \right).\]
Now do \[41(x-3)=41x-123.\]
Thus we get the final remainder as 113.
& f(x)=4{{x}^{3}}+5x-10 \\
& d(x)=(x-3) \\
& q(x)=4{{x}^{2}}+12x+41 \\
& r(x)=113. \\
Thus we got the remainder as 113, using the remainder theorem.
Option D is the correct answer.

We should know that the remainder theorem only works when a function is divided by a linear polynomial. It should be of the form (x + number) or (x - number). For e.g.:- (x + a) or (x - a), where a can be any integer.