Question

when polynomial ax cube -3x square +1 and 5x cube -ax +3 are divided by x+2, they the same remainder. find the value of a.

Answer 1

By Remainder Theorem,

x + 2 = 0 

x = - 2 


Then,



Remainder of ax^3 - 3x^2 + 1 = Remaider of 5x^3 - ax + 3 

a(-2)^3 - 3(-2)^2 + 1 = 5(-2)^3 - a(-2) + 3 

a(-8) - 3(4) + 1 = 5(-8) - a(-2) + 3 

-8a - 12 + 1 = -40 + 2a + 3 

-8a - 11 =  -37 + 2a 

-8a - 2a = -37 + 11 

-10a  = - 26 

10a = 26 

a = 26/10

a = 13/5 

Answer 2

•Here's ur Answer..

p(-2) = ax³-3x²+1
= -8a -12 +1

p(-2) = 5x³-ax +3
= -40-2a +3


✓On Keeping both the eqns ...equal..

-8a -12+1 = -40-2a+3
-6a. = -40+3-11
a. = (-51+3)/6
a = - 48/6

-Therefore the value of a is (-8..)-Answer..

#CheerS..!! ^_^