# Solve the following equation in the real number system: x^3 +x^2 -22x +8= 0

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The equation x^3 + x^2 - 22x + 8 = 0 has to be solved for real values of x.

x^3 + x^2 - 22x + 8 = 0

=> x^3 - 4x^2 + 5x^2 - 20x - 2x + 8 = 0

=> x^2(x - 4) + 5x(x - 4) - 2(x - 4) = 0

=> (x - 4)(x^2 + 5x - 2) = 0

For x - 4 = 0 , x = 4

For x^2 + 5x - 2 = 0

x1 = `(-5 + sqrt(25 + 8))/2`

=> `-5/2 + (sqrt 33)/2`

x2 = `-5/2 - (sqrt 33)/2`

**The real roots of x^3 + x^2 - 22x + 8 = 0 are x = 4, x = `-2.5 + sqrt 33/2` and x = **`-2.5 - sqrt 33/2`

You need to perform rational roots test, hence you need to list the possible rational zeroes of polynomial such that: ` {+-1;+-2;+-4;+-8}.`

Notice that you may form this set of possible zeroes creating fractions that have the following structure: all factors of constant term/all factors of leading coefficient.

Since the constant term is 8, then the factors of 8 are `{+-1;+-2;+-4;+-8}. ` Notice that the leading coefficient is 1, hence the factors of 1 are `{+-1}.`

You need to try x=-1 to check if it cancels the polynomial such that:

`(-1)^3 +(-1)^2 -22(-1) +8= 0`

`` `-1 + 1 + 22 + 8 != 0`

Hence x = -1 is not a zero for the polynomial.

You need to try `x=-2 ` to check if it cancels the polynomial such that:

`(-2)^3 +(-2)^2 -22(-2) +8= 0` `-8 + 4 + 44 + 8 != 0 =gt x=-2` is not a root for polynomial

You need to try x=4 to check if it cancels the polynomial such that:

`(4)^3 +(4)^2 -22(4) +8= 0`

`` `64 + 16 - 88 + 8 = 0 =gt x = 4` is a root for polynomial and x - 4 is a factor for the given polynomial, hence:

`x^3 +x^2 -22x +8 = (x-4)(ax^2 + bx+c)`

You need to open the brackets to the right side such that:

`x^3 +x^2 -22x +8= ax^3 + bx^2 + cx - 4ax^2 - 4bx - 4c`

Collecting like terms to the right side yields:

`x^3 +x^2 -22x +8= ax^3 + x^2(b - 4a) + x(c - 4b) - 4c`

Equating coefficients of like powers yields:

`a=1`

`b-4a=1 =gt b=5`

`4c=-8 =gt c = -2`

Hence, the quotient `ax^2 + bx+c` is `x^2 + 5x - 2 ` and if `x^3 +x^2 -22x +8 = 0` , then `(x-4)(x^2 + 5x - 2) = 0=gt x^2 + 5x - 2 = 0` .

Henc, you need to use quadratic formula to find the roots of `x^2 + 5x - 2 = 0` such that:

`x_(2,3) = (-5+-sqrt(25 + 8))/2= gt x_2 = (-5+sqrt(33))/2`

`x_3 = (-5-sqrt(33))/2`

The graph sketched below proves thatit intercepts x axis at the pooints `x_1=4, x_2 = (-5+sqrt(33))/2; x_3 = (-5-sqrt(33))/2` such that:

**Hence, the solutions to the polynomial equation are `x_1=4, x_2 = (-5+sqrt(33))/2; x_3 = (-5-sqrt(33))/2.` **