Question

Using the remainder theorem find the remainder when f (x) is divided by g (x).
\[f\left( x \right)={{x}^{24}}-{{x}^{19}}-2\]
\[g\left( x \right)=x+1\]

Answer 1

Hint: Remainder theorem is stated as “If the polynomial p (x) is divided by the binomial x – a, the remainder obtained is p (a). Comparing f (x) with p (x) and g (x) by x – a and try to get the value of ‘a’ to get the remainder, p (a) = f (a).

Complete step-by-step answer:
We are given that,
\[f\left( x \right)={{x}^{24}}-{{x}^{19}}-2\]
\[g\left( x \right)=x+1\]
The Remainder theorem is stated as “If the polynomial p (x) is divided by the binomial x – a, the remainder obtained is p (a). So comparing this with the above remainder theorem, we have,
\[f\left( x \right)=p\left( x \right)\]
f (x) is divided by the binomial \[x+1=x-\left( -1 \right)=g\left( x \right).\] The remainder obtained is \[p\left( a \right)=f\left( a \right).\]
Here,
\[g\left( x \right)=x+1\]
\[\Rightarrow g\left( x \right)=x-\left( -1 \right)\]
And comparing this with x – a, we have a = – 1.
The remainder by using the remainder theorem will be,
\[f\left( a \right)=f\left( -1 \right)={{\left( -1 \right)}^{24}}-{{\left( -1 \right)}^{19}}-2\]
\[\Rightarrow f\left( -1 \right)=1-\left( -1 \right)-2\]
\[\Rightarrow f\left( -1 \right)=1+1-2\]
\[\Rightarrow f\left( -1 \right)=0\]
So, using the remainder theorem when \[f\left( x \right)={{x}^{24}}-{{x}^{19}}-2\] is divided by \[g\left( x \right)=x+1\] is \[f\left( a \right)=0.\]
Therefore, the answer is \[f\left( a \right)=0\] is the remainder.

Note: The possibility of a mistake can be at the point where \[g\left( x \right)=x+1\] is considered. Observe that the remainder theorem, we have p (x) is divided by the binomial x – a, observe here that it is x – a and not x + a. So, while we are considering \[g\left( x \right)=x+1\] we need to write g (x) as \[x-\left( -1 \right),\] so that we can get a = – 1.