Question

find the answer pls dear

Answer 1

$<b>HELLO !$

put x-4=0

x=4

REPLACE the Value of in ax^3+3x^2-3


=a×4^3+3×4^2-3

=64a+48-3

=64a+45
So the first remainder is 64a+45.

AGAIN REPLACE the value of x =4 in 2x^3-5x+a


=2×4^3-5×4+a

=128a-20+a

=129a-20..(1)

the second remainder is 129a-20

According to question remainder are same so

129a-20=64a+45


129a-64a=45+20

65a=65

a=65/65

a=1

So, the value of a is $\red{\huge{1}}$