solve the equation x^3-5x^2-4x+20=0
I need a hint on how to solve this
Look for the first root by graphical means or trial-and-error. Note that when x=2,
x^3-5x^2-4x+20 = 8-20 -8 +20 = 0.
Therefore 2 is one root and x-2 is a factor.
Next, divide x-2 into x^3-5x^2-4x+20 using long division, to get a quadratic factor. The result is
(x-2)(x^2-3x-10)
The quadratic factor is just (x-5)(x+2), so the equation becomes
(x-2)(x+2)(x-5) = 0
This tells you that the other roots are x = -2 and +5.
To solve the equation x^3-5x^2-4x+20=0, follow these steps:
1. Graphical Method: Graph the equation on a graphing calculator or software. Look for the x-values where the graph intersects the x-axis, indicating that the equation is equal to zero at those points. In this case, when x=2, the equation evaluates to zero.
2. Trial and Error Method: Substitute different values for x into the equation and see if they make the equation equal to zero. In this case, when x=2, the equation evaluates to zero.
Since x=2 is one root of the equation, we know that (x-2) is a factor.
3. Long Division: Divide the expression (x^3-5x^2-4x+20) by (x-2) using long division to find the other factor. The quotient you get is (x^2-3x-10).
4. Quadratic Factor: Factor the quadratic equation x^2-3x-10. In this case, the quadratic factors are (x-5)(x+2).
5. Set the equation equal to zero: The equation can be written as (x-2)(x+2)(x-5) = 0.
This means the other roots of the equation are x=-2 and x=5.
So, the solutions to the equation x^3-5x^2-4x+20=0 are x=2, x=-2, and x=5.