Question

In each of the following use the remainder theorem. Find the remainder when \[f(x)\] is divided by \[g(x)\] and verify the result by actual division.
\[f(x) = 4{x^3} - 12{x^2} + 14x - 3\] , \[g(x) = 2x - 1\]

Answer 1

Hint: We first use the remainder theorem to find the remainder of \[f(x) = 4{x^3} - 12{x^2} + 14x - 3\] when divided by \[g(x) = 2x - 1\] and then using the method of long division find the remainder and check if both the remainders are equal.
* The remainder theorem states that if a polynomial \[f(x)\] is divided by a linear expression \[x - A\], then the remainder will always be the same as \[f(A)\].
* Long division method: when dividing \[a{x^n} + b{x^{n - 1}} + ....c\] by \[px + q\] we perform as
\[px + q)\overline {a{x^n} + b{x^{n - 1}} + ....c} ((a/p){x^{n - 1}} + ...\]
            \[\underline { - a{x^n} + (qa/p){x^{n - 1}}} \]
                          \[0.{x^n} + (b - qa/p){x^{n - 1}}\]
     Here we multiply the divisor with such term that gives us the exact same term as the highest power in the dividend and then we proceed in the same way.
We multiply the divisor with such a factor so we cancel out the highest power of the variable in it.

Complete step-by-step answer:
We have a polynomial \[f(x) = 4{x^3} - 12{x^2} + 14x - 3\]
and \[g(x) = 2x - 1\]
First we equate \[g(x) = 0\] which will give us a value of \[x\] .
\[2x - 1 = 0\]
Taking the constants terms only to the right side of the equation
\[
  2x = 0 + 1 \\
  2x = 1 \\
 \]
Now, we divide both sides of the equation by \[2\] to find the value of \[x\].
\[\dfrac{{2x}}{2} = \dfrac{1}{2}\]
Cancelling out same terms from both numerator and denominator on LHS of the equation.
 \[x = \dfrac{1}{2}\]
Find the remainder we substitute the value of \[x = \dfrac{1}{2}\] in the function \[f(x) = 4{x^3} - 12{x^2} + 14x - 3\] Therefore, \[f(\dfrac{1}{2}) = 4{\left( {\dfrac{1}{2}} \right)^3} - 12{\left( {\dfrac{1}{2}} \right)^2} + 14\left( {\dfrac{1}{2}} \right) - 3\]
Since we know \[{a^n} = \underbrace {a \times a \times a...... \times a}_n\]
So, \[{\left( {\dfrac{1}{2}} \right)^3} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}\] , \[{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\].
Substituting these values in the equation
\[
  f(\dfrac{1}{2}) = 4 \times \dfrac{1}{8} - 12 \times \dfrac{1}{4} + 14 \times \dfrac{1}{2} - 3 \\
  f(\dfrac{1}{2}) = \dfrac{4}{8} - \dfrac{{12}}{4} + \dfrac{{14}}{2} - 3 \\
 \]
Take LCM on RHS of the equation.
\[f(\dfrac{1}{2}) = \dfrac{{4 - 12 \times 2 + 14 \times 4 - 3 \times 8}}{8}\]
\[f(\dfrac{1}{2}) = \dfrac{{4 - 24 + 56 - 24}}{8}\]
\[f(\dfrac{1}{2}) = \dfrac{{(56 + 4) - (24 + 24)}}{8}\]
\[f(\dfrac{1}{2}) = \dfrac{{60 - 48}}{8}\]
\[f(\dfrac{1}{2}) = \dfrac{{12}}{8}\]
Cancel out same terms from numerator and denominator
\[f(\dfrac{1}{2}) = \dfrac{{4 \times 3}}{{4 \times 2}}\]
\[f(\dfrac{1}{2}) = \dfrac{3}{2}\]
Thus, from remainder theorem we have \[f(\dfrac{1}{2}) = \dfrac{3}{2}\]
Now we show the remainder by long division method using the method explained in the hint part and always multiply the divisor with a factor that cancels out the highest power of the equation in the dividend and after the highest power term becomes zero we do the same to the next highest power term.
\[2x - 1)\overline {4{x^3} - 12{x^2} + 14x - 3} (2{x^2} - 5x + \dfrac{9}{2}\]
                 \[\underline { - 4{x^3} - 2{x^2}} \]
                           \[ - 10{x^2} + 14x - 3\]
                           \[\underline { - 10{x^2} + 5x} \]
                                          \[9x - 3\]
                                        \[\underline { - 9x - \dfrac{9}{2}} \]
                                                   \[\dfrac{3}{2}\]
Therefore, by long division method the remainder comes out to be \[\dfrac{3}{2}\].
Thus, the remainder is the same when obtained by the long division method and when obtained by the remainder theorem.

Note: Students are likely to make mistakes while performing the long division method, always keep in mind that sign needs to be changed from negative to positive and vice versa inside the division when we are solving for the next value to be divided by the dividend.
Students should know the process of long division which is just like basic division just involving polynomial equations and the sign change which is the most important part.