if ax^{3}+bx^{2}+x-6 has x+2 as factor and leaves the remainder 4 when divided by (x-2), find the values of 'a' and 'b' .

it can be solved as follows :

let p(X)=ax^{3}+bx^{2}+x-6

given that p(2)=4

==>a.2^{4}+b.2^{2}+2-6=4

==>16a+4b-4=4

==>16a+b=8......(*1)*

now

p(-2)=0

==> a.-2^{3}+b.-2^{2}+(-2)-6=0

==>-8a+4b-2-6=0

==>-8a+4b-8=0

==>-8a+4b=8 (taking 4 common and dividing 8 by 4)

==>2a+b=2 ......(2)

on subtracting 1 from 2 we get,

2a+b -(16a+b)=2-8

==> 2a+b-16a-b=-6

-14a=-6

a=-6/-14 ==> a=3/7

hope the answer is correct..

on putting a in equation 1 we get,

16.3/7+b=8

b=8/7

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