Question

If $4x^4–3x^3–3x^2+x–7$ is divided by $1–2x$ then the remainder will be:

• A. $\frac{57}{8}$
• B. $-\frac{59}{8}$
• C. $\frac{55}{8}$
• D. $-\frac{55}{8}$

Answer 1

The correct option is B

$-\frac{59}{8}$

Explanation of correct option:

Finding the remainder:

Putting the value of $1–2x$ i.e. $x=\frac{1}{2}$ in $4x^4–3x^3–3x^2+x–7$, we will get the remainder.

$$\begin{gathered} =4{\left(\frac{1}{2}\right)}^4–3{\left(\frac{1}{2}\right)}^3–3{\left(\frac{1}{2}\right)}^2+\left(\frac{1}{2}\right)–7\left[\because 1-2x=0,x=\frac{1}{2}\right] \\ =4\times \frac{1}{16}-3\times \frac{1}{8}-3\times \frac{1}{4}+\frac{1}{2}-7 \\ =\frac{1}{4}-\frac{3}{8}-\frac{3}{4}+\frac{1}{2}-7 \\ =\frac{2-3-6+4-56}{8} \\ =-\frac{59}{8} \end{gathered}$$

Hence, the remainder is $-\frac{59}{8}$

Hence, option (B) is correct.