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# Find the remainder when $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$ is divide by $g\left( x \right)=2x-1$

Last updated date: 19th Jul 2024
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Hint: In this type of question we have to use the concept of remainder theorem. We know that remainder theorem states that, If a polynomial $P\left( x \right)$ is divided by the binomial $\left( x-a \right)$, then the remainder obtained is equal to $P\left( a \right)$. Here, first we find the value of $x$ by equating $g\left( x \right)$ to zero and then we find the remainder of $f\left( x \right)$ by using the remainder theorem.

Complete step-by-step solution:
Now, we have to find the remainder of $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$ when it is divided by $g\left( x \right)=2x-1$.
For this let us consider,
\begin{align} & \Rightarrow g\left( x \right)=0 \\ & \Rightarrow 2x-1=0 \\ & \Rightarrow 2x=1 \\ & \Rightarrow x=\dfrac{1}{2} \\ \end{align}
Now, as remainder theorem states that, If a polynomial $P\left( x \right)$ is divided by the binomial $\left( x-a \right)$, then the remainder obtained is equal to $P\left( a \right)$. Hence, the remainder of $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$ when it is divided by $g\left( x \right)=2x-1$ is $f\left( \dfrac{1}{2} \right)$.
Hence, now we have to find the value of $f\left( \dfrac{1}{2} \right)$.
As we have given that, $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$
\begin{align} & \Rightarrow f\left( \dfrac{1}{2} \right)=4{{\left( \dfrac{1}{2} \right)}^{3}}-12{{\left( \dfrac{1}{2} \right)}^{2}}+14\left( \dfrac{1}{2} \right)-3 \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=4\left( \dfrac{1}{8} \right)-12\left( \dfrac{1}{4} \right)+14\left( \dfrac{1}{2} \right)-3 \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{2} \right)-3+7-3 \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{2} \right)+1 \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{3}{2} \\ \end{align}
Hence, the remainder of $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$ when it is divided by $g\left( x \right)=2x-1$ is $f\left( \dfrac{1}{2} \right)=\dfrac{3}{2}$.

Note: In this type of question students have to remember the remainder theorem. Also students have to note that we can apply the remainder theorem only if the divisor is in the form $\left( x-a \right)$. One of the students may solve the question in another way as follows:
First, arrange the polynomials (dividend and divisor) in the decreasing order of its degree. Then Divide the first term of the dividend by the first term of the divisor to produce the first term of the quotient. After that, multiply the divisor by the first term of the quotient and subtract this product from the dividend, to get the remainder. This remainder is the dividend now and the divisor will remain the same. Continue the process until the degree of the new dividend is less than the degree of the divisor.
\Rightarrow 2x-1\overset{2{{x}^{2}}-5x+\dfrac{9}{2}}{\overline{\left){\begin{align} & 4{{x}^{3}}-12{{x}^{2}}+14x-3 \\ & - \\ & 4{{x}^{3}}-2{{x}^{2}} \\ & \overline{\begin{align} & 00-10{{x}^{2}}+14x-3 \\ & \,\,\,\,\,\,\,\,-\left(-10{{x}^{2}}+5x\right) \\ & \overline{\begin{align} & 9x-3 \\ & \underline{- \left( 9x-\dfrac{9}{2} \right)}\\ {\dfrac{3}{2}} \\ \end{align}} \\ \end{align}} \\ \end{align}}\right.}}
Hence, the remainder of $f\left( x \right)=4{{x}^{3}}-12{{x}^{2}}+14x-3$ when it is divided by $g\left( x \right)=2x-1$ is $f\left( \dfrac{1}{2} \right)=\dfrac{3}{2}$.