Find the values of `a\ a n d\ b` so that `(x+1)a n d\ (x-1)` are factors of `x^4+a x^3-3x^2+2x+b`

Explanation

1409718 Let `p(x)=x^4+ax^3-3x^2+2x+b`
If `(x+1)` and `(x-1)` are factors of `p(x)` then by factor theorem:
`=>x+1=0`
`=>x=-1`
`p(-1)=0`
`=>x-1=0`
`=>x=1`
`p(1)=0`
Now,
`p(-1)=0`
`=>(-1)^4 +a(-1)^3-3(-1)^2+2(-1)+b=0`
`=>1-a-3-2+b=0`
`=>-a+b=4` ...(1)
and `p(1)=0`
`=>(1)^4+a(1)^3-3(1)^2+2(1)+b=0`
`=>1+a-3+2+b=0`
`=>a+b=0` ...(2)
Solving (1) and (2),we get:
`a=-2` and `b=2`.