It is given that $3x^3-21x^2-22x+24$ when divided by $x+p$ has a remainder that is half the

Question

It is given that $3x^3-21x^2-22x+24$ when divided by $x+p$ has a remainder that is half the remainder when it is divided by $x-p$. Show that $3p^3+7p^2-22p-8=0$.[3] Solve $3p^3+7p^2-22p-8=0$.[6] Hence solve $3p^3-7p^2-22p+8=0$.[2]

Accepted Answer

(i) $3p^3+7p^2-22p-8=0$ (ii) $p=-4,-\dfrac13,2$ (iii) $p=-2,\dfrac13,4$

Polynomials, Cubic Equations and Partial Fractions - Remainder and Factor Theorems

Polynomials, Cubic Equations and Partial Fractions - Remainder and Factor Theorems

Video Solution:

Solution:

Solution locked