# Solve the following equation in the real number system. 3x^3 -20x^2 +29x +12 =0

*print*Print*list*Cite

### 1 Answer

You should perform the rational root test, hence you nional reed to form a set of possible rational roots of equation.

You need to remember that the numerator of these possible rational roots is one of the factors of constant term 12 and the denominator is one of the factors of leading coefficient 3, hence the possible rational roots are `{+-1/3;+-2/3;+-2;+-3;+-4;+-4/3;+-6;+-12} ` .

You need to try the first rational value `x = -1/3` :

`-3/27-20/9 - 29/3 + 12 =0`

`` `-7/3 - 29/3 + 12 = 0`

Bringing all terms to a common denominator yields:

`(-36 + 3*12)/3 = 0 =gt 0=0`

Hence, since `x=-1/3` cancels the equation, then you may say that you found a root.

You should apply the reminder theorem to determine the quotient, that is a second order polynomial such that:

`3x^3 -20x^2 +29x +12 = (x+1/3)(ax^2 + bx + c)`

Opening the brackets to the right yields:

`3x^3 -20x^2 +29x +12 = ax^3 + bx^2 + cx + ax^2/3 + bx/3 + c/3`

Collecting like terms yields:

`3x^3 -20x^2 +29x +12 = ax^3 + x^2(b + a/3) + x(c + b/3) + c/3`

Equating the coefficients of like terms both sides yields:

`a=3`

`` `b + a/3 = -20 =gt b + 1 = -20 =gt b = -21 `

`c/3 = 12 =gt c = 36`

Hence, the quotient is 3x^2 -21x + 36.

Since `3x^3 -20x^2 +29x +12 =0 =gt (x+1/3)(3x^2- 21x + 36) = 0`

You need to find the roots of quotient `3x^2 -21x + 36 = 0` , hence you need to divide by 3 and you need to use quadratic formula:

`x^2 - 7x + 12 = 0`

`` `x_(1,2) = (7+-sqrt(49 - 48))/2 =gt x_(1,2) = (7+-1)/2`

`x_1 = 4; x_2 = 3`

Hence,the graph of the function intercepts x axis at x=-1/3, x=3 and x=4 as the sktech below proves:

**Hence, the real roots to the equation are selected from the set of possible rational roots and they are `x_1=-1/3, x_2=3;x_3=4.` **