1.

The value of a for which `x^3-7x+5` is a factor of `x^5-2x^4-4x^3+19x^2-31x+21+a`

Answer» Here, `f(x) = x^5-2x^4-4x^3+19x^2-31x+21+a`
`g(x) = x^3-7x+5`
For `g(x)` to be a factor of `f(x)`, `f(x)` should be completely divisible by `0`.
`:.` Remainder of `f(x)/g(x)` should be `0`.
If we divide `f(x)` by `g(x)`, we get,
`x^5-2x^4-4x^3+19x^2-31x+21+a = (x^3-7x+5)(x^2-2x+3) +(6+a)`
As remainder is `0`,
`:. 6+a = 0=> a = -6`


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