Question

Question #4a732

Answer 1

If we divide `[x^3-2x^2+ax-b]` by `[x^2-2x-3]`, there is a remainder ( x - 6 ). It means `[x^3-2x^2+ax-b-(x-6)]` is fully divisible by `[x^2-2x-3]`.

Henceforth, `[x^2-2x-3]` go 'x' times = `x^3-2x^2-3x` If we subtract this from `[x^3-2x^2+(a-1)x-(b-6)]` it becomes `(a-1)x+3x-(b-6)`

Therefore, `(a-1)x = 3x & -(b-6)=0`

`rArr a = 4 & b = 6`

Answer 2

`a=-2, and, b=6.`

Explanation:

Suppose that, when `p(x)=x^3-2x^2+ax-b` is divided by

`f(x)=x^2-2x-3,` the Quotient Poly. is `q(x)` and, the

Remainder Poly. is `r(x)=x-6.`

Since, Dividend=QuotientxxDivisor+Remainder, we have,

`p(x)=f(x)xxq(x)+r(x), or,`

`x^3-2x^2+ax-b=(x^2-2x-3)q(x)+(x-6)...........(ast).`

`because, (ast)" holds good "AA x in RR," we have,"`

`x=-1 rArr -1-2-a-b=(1+2-3)q(x)-7," giving, "`

`a+b=4.....................(1).`

`x=3 rArr 27-18+3a-b=(9-6-3)q(x)-3," so that, "`

`3a-b=-12.....................(2).`

Solving, `(1) and (2),` we get,

`a=-2, and, b=6.`

Enjoy Maths.!