How do you use the remainder theorem to evaluate \[f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7\] at \[a=3\]?

Question

How do you use the remainder theorem to evaluate $$f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7$$ at $$a=3$$?

Vedantu Content Team · Accepted Answer

Hint: We are given a polynomial expression and we have to find its value at $$a=3$$ using the remainder theorem. Remainder theorem states that the if a polynomial $$f(x)$$ is divided by $$x-a$$ then the remainder is $$f(a)$$.  We will use the long division method for dividing $$f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7$$ by $$a-3$$ and then we will have the value of the polynomial expression at $$a=3$$ as the remainder of our long division. Complete step-by-step answer:According to the given question, we are given a polynomial equation and we have to find the value of this equation at $$a=3$$ using the remainder theorem. Remainder theorem states that, if a polynomial $$f(x)$$ is divided by $$x-a$$ then the remainder is $$f(a)$$. That is, the remainder we will get on dividing is equivalent to the value we will get for $$f(x)$$ when ‘x’ is substituted as ‘a’. The polynomial equation we have is,$$f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7$$We have to evaluate the given equation at $$a=3$$. Since, we have to proceed using the remainder theorem, we can $$a=3$$ as $$a-3$$.We will use the long division method here, so we get,$$a-3\overset{{{a}^{3}}+6{{a}^{2}}+a+5}{\overline{\left){\begin{align}  & {{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7 \\  & \underline{-({{a}^{4}}-3{{a}^{3}})} \\  & 0{{a}^{4}}+6{{a}^{3}}-17{{a}^{2}}+2a-7 \\  & \underline{-(6{{a}^{3}}-18{{a}^{2}})} \\  & 0{{a}^{3}}+{{a}^{2}}+2a-7 \\  & \underline{-({{a}^{2}}-3a)} \\  & 0{{a}^{2}}+5a-7 \\  & \underline{-(5a-15)} \\  & \_\_\_\_8\_\_ \\ \end{align}}\right.}}$$     We have done the long division as per required and we have got a quotient and a remainder.The quotient we have is, $${{a}^{3}}+6{{a}^{2}}+a+5$$ And the remainder we got is $$8$$.Therefore, the value of the given polynomial equation at $$a=3$$ is 8.Note: The remainder we got after doing the long division method is equivalent to the value we will on substituting $$a=3$$ in $$f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7$$ and which is also 8. So, we can verify if our obtained answer is correct or not. The long division should be done carefully without any mistakes.