This is an excellent question.
X3 + X2 - X = 25
Step 1: X3 + X2 - X - 25 = 0
Step 2: Plug this in to your calculator's graph. Find the roots (in other words, where y= 0 or intersects the line y=0).
Step 3: Answer is approximately 2.73
You can't do it any faster. Another way of solving this problem is by using the quadratic formula for cubic functions but that is a pain and completely unnecessary when your teacher allows the calculator. You cannot factor the problem out either. This leaves you with ths option.
Solving general general cubic and quartic polynomials by radicals is a problem with an interesting history.
Anyways, if you want to solve your numerically, you can use Newton's method to get a real root:
x ≈ 2.72741
Factoring x - 2.72741... out of your cubic x3 + x2 - x - 25 = 0 gives a quadratic, which when set to 0 and solved by quadratic formula gives the other two roots, which are complex:
x ≈ -1.86371 ± 2.38596 i
If yo want to solve your cubic by radicals, you can do it as follows, since it fails the rational root test.
In the cubic ax3+bx2+cx+d=0, if you substitute x = y - b/(3a), expand and simplify by collecting terms with equal powers of y, you get a cubic polynomial in y in which there is no y2 term. We also need to divide both sides by a and put the linear and constant term on the right hand side. The resulting cubic is called a depressed cubic (general form is y3 = py + q).
With your cubic, x3 + x2 - x - 25 = 0, we let x = y - 1/(3*1) = y - 1/3 and substitute.
(y - 1/3)3 + (y - 1/3)2 - (y - 1/3) - 25 = 0
y3 - y2 + (1/3)y - 1/27 + y2 - (2/3)y + 1/9 - y + 1/3 - 25 = 0
y3 - (4/3)y - 664/27 = 0
y3 = (4/3)y + 664/27 (This is our depressed cubic equation).
Step 2: Solving the depressed cubic.
Observe the identity (s+t)3 = 3st(s+t) + s3+t3. I will leave it to you as an exercise to verify this identity.
Notice that it is of the form y3 = py + q where p = 3st, q = s3 + t3, and y =s+t.
Thus if we find two numbers p = 3st, q = s3 + t3, we will find that y = s + t is a root.
If we factor out y-(s+t) and solve the resulting quadratic by the quadratic formula, it turns out that the other two roots are
y = -(s+t)/2 ± (√3)(s-t)i/2.
With your cubic, for example, we derived the depressed cubic and saw that p = 4/3 and q = 664/27 so we seek numbers s and t satisfying 3st = 4/3 and s3 + t3 = 664/27.
In the first equation, dividing by 3 and cubing gives s3t3 = 64/729. Thus s3 and t3 are roots of a quadratic
u2 - (664/27)u +64/729 = 0
which when solved using the quadratic formula gives u = (332 ± 36√85)/27.
Taking cube roots give the expressions for s and t so we can back substitute in the formulas for y and then back substitute the y values to get x. Here is the final result:
x = (-1 + 3√(332 + 36√85) + 3√(332 - 36√85) )/3
x = -1/3 - (3√(332 + 36√85) + 3√(332 - 36√85))/6 + i(√3)(3√(332 + 36√85) - 3√(332 - 36√85))/6
x = -1/3 - (3√(332 + 36√85) + 3√(332 - 36√85))/6 - i(√3)(3√(332 + 36√85) - 3√(332 - 36√85))/6
As noted above, this will not factor. However in a 3rd degree polynomial there are 3 roots.
The real root is
The two complex roots are
-1.8637 ± 2.3860i