Question
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by:$5 + 2x$
✓
Answer
Here, $f(x) = x^3 + 3x^2 + 3x + 1$ By remainder theorem $5 + 2x = 0 2x = -5$$\text{x}=\frac{-5}{2}$
substitute the value of x in f(x)$\text{f}\Big(\frac{-5}{2}\Big)=\Big(\frac{-5}{2}\Big)^3+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+3\Big(\frac{25}{4}\Big)+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$
Taking L.C.M$=\frac{-125+150-50+8}{8}$
$=\frac{-27}{8}$
substitute the value of x in f(x)$\text{f}\Big(\frac{-5}{2}\Big)=\Big(\frac{-5}{2}\Big)^3+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+3\Big(\frac{25}{4}\Big)+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$
Taking L.C.M$=\frac{-125+150-50+8}{8}$
$=\frac{-27}{8}$
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
In the adjoining figure, CE || AD and CF || BA. Prove that $\text{ar}(\triangle\text{CBG})=\text{ar}(\triangle\text{AFG}).$
→It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following:$\frac{1+2\sqrt{3}}{2-\sqrt{3}}$
→If the median of scores $\frac{\text{x}}2{},\frac{\text{x}}3{},\frac{\text{x}}4{},\frac{\text{x}}5{}$ and $\frac{\text{x}}6{}$ (where x > 0) is 6, then find the value of $\frac{\text{x}}6{}.$
→If $\frac{15 a^2+4 b^2}{15 a^2-4 b^2}=\frac{47}{7}$, then find the value of the following ratios : $\frac{b^2-2 a^2}{b^2+2 a^2}$
→Factorize the following expressions:$8x^3 - 125y^3 + 180xy + 216$
→In the given figure, y = 108° and x = 71°. Are the lines m and n parallel? Justify?
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: $\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CPD})\\=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC})$
→In the given figure, BE and CF are two equal altitudes of $\triangle\text{ABC}.$
Show that:
→Show that:
- $\triangle\text{ABE}\cong\triangle\text{ACF},$
- AB = AC.
The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of ₹ 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.
Given: O is the centre of the circle.
diameter CD ⊥ chord AB, A-E-B
To prove: ∆ABC is an isosceles triangle.