Question

Find the remainder when \[f\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by \[g\left( x \right)=1-2x\]

Answer 1

Hint: According to the remainder theorem, if a polynomial f(x) is divided by a factor \[\left( x-a \right)\], then f(a) gives the remainder. To find the remainder of the given division, apply the remainder theorem. First set \[g\left( x \right)=0\], and find the value of x. Then substitute this value of x in f(x), and simplify the expression. The resulting value will be the remainder. Use rules of BODMAS and addition and subtraction of integers for simplification. Find LCM to add and subtract fractions.

Complete step-by-step answer:
First set \[g\left( x \right)=0\], and find the value of x. Then substitute this value of x in f(x), and simplify the expression. The resulting value will be the remainder. Use rules of BODMAS and addition and subtraction of integers for simplification. Find LCM to add and subtract fractions.
Given,
      \[g\left( x \right)=1-2x\]
Setting \[g\left( x \right)=0\], we get,
     \[\begin{align}
  & g\left( x \right)=1-2x \\
 & 0=1-2x \\
 & 2x=1 \\
 & x=\dfrac{1}{2} \\
\end{align}\]
Substituting this value of x in f(x), we get,
     \[\begin{align}
  & f\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4 \\
 & f\left( \dfrac{1}{2} \right)={{\left( \dfrac{1}{2} \right)}^{3}}-6{{\left( \dfrac{1}{2} \right)}^{2}}+2\left( \dfrac{1}{2} \right)-4 \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{1}{8}-6\left( \dfrac{1}{4} \right)+2\left( \dfrac{1}{2} \right)-4 \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{1}{8}-\dfrac{3}{2}+1-4 \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{1}{8}-\dfrac{3}{2}-3 \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{1-12-24}{8} \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{1-36}{8} \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{-35}{8} \\
\end{align}\]
Therefore, when \[f\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\]is divided by \[g\left( x \right)=1-2x\], remainder is \[-\dfrac{35}{8}\]

Note: Another way of doing this sum is through long division of f(x) by g(x). One other way is through synthetic division. Synthetic division is a shortcut way of dividing the polynomials without involving the variables. Remainder theorems and synthetic division methods are easier than long division methods. If the remainder was zero, g(x) would have been the factor of f(x). Hence, the remainder theorem is also used to verify factors of a polynomial. LCM is calculated using the prime factorization or division method. The answer can be verified by multiplying the divisor with the quotient and adding remainder. The resulting expression will be equal to the dividend. Although, quotients will be obtained through a long division method.