# Does the equation x^3-4x^2-x+4 =0 have a real root?

Asked on by capitol

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation x^3 - 4x^2 - x + 4 = 0 is a cubic equation with real coefficients for all the powers of x. This equation has 3 roots of which only two can be imaginary. It is not possible for a cubic root to have 3 imaginary roots. For the given equation:

x^3 - 4x^2 - x + 4 = 0

=> x^2(x - 4) - 1(x - 4) = 0

=> (x^2 - 1)(x - 4) = 0

=> (x - 1)(x + 1)(x - 4) = 0

=> x = 1, x = -1 and x = 4

The equation x^3 - 4x^2 - x + 4 = 0 has real roots. Here, they are {-1, 1, 4}

rachellopez | Student, Grade 12 | (Level 1) Valedictorian

Posted on

a root is another fancy word for solution. To find the roots of x^3-4x^2-x+4 =0, you need to factor it. Start by taking out something that they have in common to shorten it a bit.

x^3-4x^2-x+4 =0

x^2(x-4)-1(x-4)=0

combine like terms

(x^2-1)(x-4)=0

(x-1)(x+1)(x-4)

set each factor equal to zero

x-1=0              x-4=0         x+1=0

x=1                 x=4            x=-1

You have three real roots: 1, -1, 4