if ax3+bx2+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)=ax3+bx2+x-6
given that p(2)=4
==>a.24+b.22+2-6=4
==>16a+4b-4=4
==>16a+b=8......(1)
now
p(-2)=0
==> a.-23+b.-22+(-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