Factorise: $t^{4} + t^{3} - 2 t - 2$

Open in App

Solution

The equation $t^{4} + t^{3} - 2 t - 2$ can be factorised as follows:

$t^{4} + t^{3} - 2 t - 2 = (t^{4} + t^{3}) + (- 2 t - 2) = t^{3} (t + 1) - 2 (t + 1) = (t^{3} - 2) (t + 1)$

Hence, $t^{4} + t^{3} - 2 t - 2 = (t^{3} - 2) (t + 1)$

Suggest Corrections

Similar questions

Let $θ \in (0, \frac{π}{4}) a n d t_{1} = (t a n θ)^{t a n θ}, t_{2} (t a n θ)^{c o t} t_{3} = (c o t θ)^{t a n θ} a n d t_{4} (c o t θ)^{t a n θ}$ , then

Q. A ring is rotating with an angular acceleration given by

α = 4 t^{3} - \frac{3}{2} t^{2}

, where

t

is time. Find the equation of angular speed

ω

if the ring starts from rest.

Q. In an

A . P

a = 2

and

t_{5} = 12

, find

t_{2}, t_{3}, t_{4}

Q. If

\begin{matrix} T_{5} = 81 T_{2} = 3 \end{matrix}

find

T_{3}

and

T_{4}

Q. If the speed of the particle is

v = 4 t^{3} - 3 t^{2} + 3 m/s

, the distance of the particle as a function of time

t

and the average speed in

2 sec

respectively are

Join BYJU'S Learning Program

Factorise: t4+t3−2t−2

The equation t4+t3−2t−2 can be factorised as follows:t4+t3−2t−2=(t4+t3)+(−2t−2)=t3(t+1)−2(t+1)=(t3−2)(t+1)Hence, t4+t3−2t−2=(t3−2)(t+1)

Factorise: $t^{4} + t^{3} - 2 t - 2$

The equation $t^{4} + t^{3} - 2 t - 2$ can be factorised as follows:

$t^{4} + t^{3} - 2 t - 2 = (t^{4} + t^{3}) + (- 2 t - 2) = t^{3} (t + 1) - 2 (t + 1) = (t^{3} - 2) (t + 1)$

Hence, $t^{4} + t^{3} - 2 t - 2 = (t^{3} - 2) (t + 1)$