Find all real solutions to this equation?

Question

x^3 - 2x^2 - 5X + 10 = 0

Not sure how to solve this. How do you do it and what's the answer? Thanks!

Answer 1

Just as there is the 'quadratic' formula for solving quadratics, there are 'cubic' formulas, but they are more difficult to work with.

Another approach is to solve this numerically. If you have Microsoft Excel (or other spreadsheet), start a column down 'A' with numbers starting at -10, then -9, -8... all the way up to +10. In column B write a formula:
=A1^3-2*A1^2-5*A1+10
Then copy the formula down the column.
Then chart columns A and B using the X-Y Scatter type chart.

Where does this function cross the X-axis?
It looks to me like it crosses at approximately -2.5, and at +2.

So you can re-write your numbers in Column A, by 0.1. Start at -3.0, -2.9,-2.8... all the way up to +2.5. The chart should re-plot and you can get a closer look at where the x-axis crossings are. Keep refining the numbers in the A column (0.01 increments, 0.001, 0.0001, etc.) until you get more accuracy of the x value when the function crosses the x-axis.

I get:-2.2360675, 2, and 2.2360679

There may be a symmetrical solution for the 1st and last numbers but I didn't investigate any further -- you could see if a single number works for positive and negative. The +2 is an exact solution, though

OR
Use a plotting calculator (if you have one)

.

Answer 2

x^3 - 2x^2 - 5X + 10 = 0
=> x^2(x-2) -5(x-2)=0
=> (x^2-5)(x-2) = 0
=> x^2-5 = 0 OR x-2 = 0
=> x = +/- √5 OR x=2

Answer 3

x^3 - 2x^2 - 5X + 10 = 0

(x - a)(x -b)(x - c)
= (x^2 - (a +b)x + ab)(x - c)
= x^3 - (a + b +c)x^2 - (ab + bc + ca)x - abc

a + b + c = -2
ab + bc + ca = -5
abc = - 10

(x +1) (x + 2) (x - 5) = 0

x = -1, -2, 5