Question

on dividing ax^3+ 9x^2+ 4x- 10 by x+3 we get 5as a remainder .Find the value of a

Answer 1

Answer:

2

Step-by-step explanation:

X+3=0

x=-3

f(-3) =5

f(-3)=a×-3^3 + 9×-3^2 + 4x-3 -10

=-27a+81-12-10 =5

-27a= -54

a=2

hope this helps

Answer 2

Given:

$\sf rem(a{x}^{3} + 9 {x}^{2} + 4x - 10,x + 3) = 5$

So,

By remainder theorem:

$\sf a( - 3) ^{3} + 9( - 3) {}^{2} + 4( - 3) - 10 = 5$

$\sf - 27a + 81 - 12 - 10 = 5$

$\sf - 27a + 54 = 0$

$\implies \bf \underline{a = 2}$