Question

How do you solve `x^4-10x^3+27x^2-2x-40>=0`?

Answer 1

`p(x) ge 0` for `(x le -1) and (2 le x le 4) and (5 < x)`

Explanation:

Analyzing the constant term which is the product of all roots we get

`40=1cdot2^3 cdot 5`. Trying for `pm1,pm2,pm4,pm5` we get

`p(x)=x^4 - 10 x^3 + 27 x^2 - 2 x - 40`
`p(x)=(x+1)(x-2)(x-4)(x-5) ge 0`

Ordering the roots

`-1,2,4,5`

and knowing that for `x < -1->p(x)>0` we have by continuity

`-1 < x<2->p(x)<0`
`2 < x < 4 ->p(x) >0`
`4 < x < 5->p(x) < 0`
`5 < x ->p(x) > 0`

so

`p(x) ge 0` for `(x le -1) and (2 le x le 4) and (5 < x)`

Answer 2

Solution set for `x^4-10x^3+27x^2-2x-40>=` is `x<=-1` and `2 <= x <= 4` and `5 < x`.

Explanation:

Let us first factorize `f(x)=x^4-10x^3+27x^2-2x-40`. As `-1` is a zero of the function as

`f(0)=(-1)^4-10(-1)^3+27(-1)^2-2(-1)-40=1+10+27+2-40=0`, `(x+1)` is a factor of `f(x)`.

Also `f(2)=2^4-10*2^3+27*2^2-2*2-40=16-80+108-4-40=0`, hence `(x-2)` to is a factor of `f(x)`.

Dividing `f(x))=x^4-10x^3+27x^2-2x-40` by `(x+1)` and then by `(x-2)`, we get `x^2-9x+20` which can be further factorized to `(x-4)(x-5)` ad hence, the inequality we have is

`(x+1)(x-2)(x-4)(x-5)>=0`

Now, these zeros divide real number line in five parts

(1) `x<-1` - Here all the terms are negative, hence `f(x)` is positive. This forms part of solution.

(2) `-1 < x < 2` - Here while first term `(x+1)` is positive, all other terms are negative and hence `f(x)` is negative. Hence, this does not form part of solution.

(3) `2 < x < 4` - Here while first two terms `(x+1)` and `(x-2)` are positive, other two terms are negative and hence `f(x)` is positive. Hence, this forms part of solution.

(4) `4 < x < 5` - Here while first three term are positive, the last term `(x-5)` is negative and hence `f(x)` is negative. Hence, this does not form part of solution.

(5) `5 < x` - Here all the terms are positive and hence `f(x)` is positive. Hence, this forms part of solution.

Hence solution set for `x^4-10x^3+27x^2-2x-40>=` is `x<=-1` and `2 <= x <= 4` and `5 < x`.

graph{x^4-10x^3+27x^2-2x-40 [-10, 10, -40, 40]}