How to find the roots of a cubic polynomial?

Question

All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what if you can't do that?

For example, how would you find the roots of the following equation:

$x^3+7x^2+16x+12=0$

I know the roots are $-2$, $-2$ and $-3$ but don't know how to get them. If I somehow guess them, is there a way to tell which one is the double root?

Answer 1

Start out by checking the positive and negative factors of 12. Once you find one factor that makes the polynomial equal to zero, say $x = -2$, divide the polynomial by the corresponding factor $x+2$. You can use synthetic division or long division. Once you do that you get the quadratic factor $x^2+5x+6$. Factor the quadratic to get $(x+2)(x+3)$, therefore:$$x^3+7x^2+16x+12=(x+2)(x+2)(x+3)=0$$ This means $x = -2$, $-2$, and $-3$ as you pointed out but now you can see clearly that one factor repeats.

Answer 2

Factorize as follows $$x^3+7x^2+16x+12\\ =(x^3+7x^2+12x)+(4x+12)\\ =x(x+4)(x+3)+4(x+3)\\ =(x+3)(x^2+4x+4) $$

Answer 3

While there is a rather nice test for whether a cubic polynomial has a triple zero by inspecting its coefficients (it will even tell us directly the value of that root), there is not anywhere near as simple a method for testing for a double zero by using the coefficients. (It can be managed, but the calculations do not just require basic arithmetic.)

We can produce a straightforward method of testing with the coefficients if we suspect that a double zero is present. We would write for the polynomial $ \ ax^3 + bx^2 + cx + d \ = \ a · (x - r)^2 · (x - s) \ \ , $ with $ \ r \ $ being the double zero and $ \ s \ $ being the other (real) zero. This alone is not especially helpful, but we could express the "singleton" zero as $ \ s = \sigma · r \ \ . $ This then gives us $$ a · (x - r)^2 · (x \ - \ \sigma · r) \ \ = \ \ a · ( x^3 \ - \ [ \ 2r \ + \ \sigma · r \ ]· x^2 \ + \ [ \ r^2 \ + \ 2·\sigma·r^2 \ ]·x \ - \ \sigma·r^3 \ \ . $$

The relation among the coefficients will thus be $$ \frac{b}{a} \ = \ -r · (\sigma \ + \ 2) \ \ , \ \ \frac{c}{a} \ = \ r^2 · (1 \ + \ 2·\sigma) \ \ , \ \ \frac{d}{a} \ = \ -\sigma · r^3 \ \ . $$

This can be helpful for integer and "reasonably simple" rational coefficients. For your example polynomial $ \ x^3 + 7x^2 + 16x + 12 \ \ , $ the fact that the constant coefficient can be factored as $ \ 12 = 2^2 · 3 \ $ might lead us to check for a double zero. We could have either $ \ \frac{d}{a} \ = \ 12 \ = \ - \ 2^2 · (-3) \ \ $ or $ \ -(-2)^2 · (-3) \ \ . $ The Rule of Signs will tell us which of the two factorizations would be the one to try, but we will see that this test also leaves no sign ambiguity:

$$ \mathbf{r = 2 \ , \ \sigma = -\frac32 \ \ :} \quad \frac{b}{a} \ = \ -2 · \left(-\frac32 \ + \ 2 \right) \ = \ -1 \ \ , \ \ \frac{c}{a} \ = \ 2^2 · \left(1 \ + \ 2·\left[-\frac32 \right] \right) \ = \ -8 \ \ ; $$ $$ \mathbf{r = -2 \ , \ \sigma = \frac32 \ \ :} \quad \frac{b}{a} \ = \ 2 · \left(\frac32 \ + \ 2 \right) \ = \ 7 \ \ , \ \ \frac{c}{a} \ = \ (-2)^2 · \left(1 \ + \ 2·\frac32 \right) \ = \ 16 \ \ . $$

Hence, the zeroes are $ \ x \ = \ -2 \ [\text{multiplicity} \ 2] \ , \ -3 \ \ $ or $ \ x^3 + 7x^2 + 16x + 12 \ = \ (x + 2)^2 · (x \ + \ \frac32 · 2) \ = \ (x + 2)^2 · (x + 3) \ \ . $

As an additional demonstration, we could attempt to solve in this manner $ \ 9x^3 + 3x^2 - 56x + 48 \ = \ 0 \ : $

$$ \frac{d}{a} \ = \ \frac{48}{9} \ = \ \frac{16}{3} \ = \ - \left(\frac43 \right)^2 · (-3) \ \ \ [?] $$ (the Rule of Signs tells us there would be one negative zero and two positive real zeroes) so we will want to check $ \ r = \frac43 \ , \ \sigma = -\frac94 \ \ : $

$$ \rightarrow \ \ \frac{b}{a} \ = \ -\frac43 · \left(-\frac94 \ + \ 2 \right) \ = \ \frac13 \ \ , \ \ \frac{c}{a} \ = \ \left(\frac43 \right)^2 · \left(1 \ + \ 2·\left[-\frac94 \right] \right) \ = \ -\frac{56}{9} \ \ . $$

We do have $ \ b = \frac93 = 3 \ $ and $ \ c = -56 \ \ , $ which permits us to conclude that $$ 9x^3 + 3x^2 - 56x + 48 \ \ = \ \ 9 · \left( x - \frac43 \right)^2 · \ (x + 3 ) \ \ . $$

Naturally, if this coefficient test indicates that we do not have a double zero, we will need to "fall back" to more elaborate methods to determine the zeroes.