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
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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). \[f(x)=d(x).q(x)+r(x)\] 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. \[\begin{align} & f(x)=4{{x}^{3}}+5x-10 \\ & d(x)=(x-3) \\ & q(x)=4{{x}^{2}}+12x+41 \\ & r(x)=113. \\ \end{align}\] Thus we got the remainder as 113, using the remainder theorem. Option D is the correct answer.
Note: 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.