Divide $2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$ by $t^{2}-3$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The dividend is $p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$ and the divisor is $s(t) = t^2 - 3$.
Step $1$: Divide the first term of the dividend $(2t^4)$ by the first term of the divisor $(t^2)$ to get $2t^2$. Multiply $2t^2$ by $(t^2 - 3)$ to get $2t^4 - 6t^2$. Subtract this from the dividend.
Step $2$: The result is $3t^3 + 4t^2 - 9t - 12$. Divide $3t^3$ by $t^2$ to get $3t$. Multiply $3t$ by $(t^2 - 3)$ to get $3t^3 - 9t$. Subtract this from the current expression.
Step $3$: The result is $4t^2 - 12$. Divide $4t^2$ by $t^2$ to get $4$. Multiply $4$ by $(t^2 - 3)$ to get $4t^2 - 12$. Subtract this to get a remainder of $0$.
Thus,the quotient is $2t^2 + 3t + 4$ and the remainder is $0$.

Explore More

Similar Questions

For the quadratic polynomial $p(x) = ax^2 + bx + c$; if $a = 6$,$b = 11$,and $c = 4$,then the quadratic polynomial is..........

$p(x) = 3x^2 + 7x + 4$ is a $\dots$ polynomial.

If the zero of the polynomial $p(x) = ax^2 - 11x + 3$ is $1$,then $a = \ldots$

Prove that $1/2, 1$ and $-2$ are the zeros of the cubic polynomial $p(x) = 2x^3 + x^2 - 5x + 2$. Also,verify the relationship between the zeros and the coefficients.

If the polynomial $p(x) = x^{3} - 3x^{2} + x + 2$ is divided by the divisor polynomial $s(x)$, then the quotient polynomial $q(x) = x - 2$ and the remainder polynomial $r(x) = -2x + 4$ are obtained. Find the divisor polynomial $s(x)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo