MCQ
When $p(x) = x^4 + 2x^3 - 3x^2 + x - 1$ is divided by $(x - 2),$ the remainder is:
- ✓$21$
- B$0$
- C$-1$
- D$-15$
✓
Answer
Correct option: A.
$21$
$x^4 + 2x^3 - 3x^2 + x - 1$
Using remainder theorem,
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 + 16 - 12 + 2 - 1$
$= 34 - 13$
$= 21$
Using remainder theorem,
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 + 16 - 12 + 2 - 1$
$= 34 - 13$
$= 21$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\sqrt2=1.4142,$ then $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}$ is equal to:
→In the given figure, $O$ is the centre of a circle in which $\angle\text{OBA}=20^\circ$ and $\angle\text{OCA}=30^\circ.$ Then, $\angle\text{BOC}=?$
→Number of circles that can be drawn through three non-collinear points is:
The value of $\{2-3(2-3)^3\}^3,$ is:
→Which one of the following is a polynomial?
The points $(-5, 3)$ and $(3, -5)$ lie in the.
→In the given figure, $\angle\text{BAC}=40^\circ,\angle\text{ACB}=90^\circ$ and $\angle\text{BED}=100^\circ$ Then $\angle\text{BDE}=?$
→The decimal expansion of $\sqrt{2}$ is:
→Write the correct answer in the following: $\sqrt{2}$ is a polynomial of degree.
→The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point $O$. If $\angle\text{DAC}=32^\circ$ and $\angle\text{AOB}=70^\circ$ then, $\angle\text{DBC}$ is equal to:
→