
How do you use the remainder theorem to evaluate \[f(a)={{a}^{4}}+3{{a}^{3}}-17{{a}^{2}}+2a-7\] at \[a=3\]?
Answer
536.1k+ views
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.
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.
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