Every cubic with real coefficients has either one real root or three. Alternatively, it has 2 complex roots or none.

In the real world, cubic equations will not be so easy to solve as they are on this page, because the roots may not be easy integers.

+ + + -

Therefore there is one positive real root of the equation. We know every cubic has at least one real root, and now we know that for this cubic, the root is positive.

As every cubic has 3 roots, then, we know immediately that this one has 2 negative real roots or none.

Just for the record, the signs of the coefficients for f(-x) are:

- + - -

Showing two sign changes and indicating 2 or no negative real roots.

This means any rational roots would be: ±(1/1, 2/1, 3/1, or 6/1).

In this case, we can quickly check that 1 is a root. Because we must use all the factors in our roots, if one factor is 1, then the other two must be ±2 or ±3, or ±1 and ±6, if the roots are rational.

We might at this stage realise that the equation can be easily factorised.

αβγ=6

With one positive root, the possible roots could be -2, -3 and 1, by inspecting the possibilities from the rational root theorem.

Other possibilities, such as 1, 1, -6 do not fit as Descartes has said there is only one positive real root. Naturally, 1 is easy to check and we find that f(1)=1+4+1-6=0, so 1 is indeed a root.

1 | 4 | 1 | -6 | |

1 | 0 | 1 | 5 | 6 |

1 | 5 | 6 | 0 | |

-2 | 0 | -2 | -6 | |

1 | 3 | 0 |

The last line is 1,3, or x+3=0, so we see the last root is indeed -3. Alternatively, after dividing by 1, we find that the result of division by x+1 is 1,5,6, or x

The original equation,

x

+ - + -

So there are 3 or 1 real positive roots. There can't be any real negative roots, as cubics have only 3 roots. However, these are the signs for f(-x):

- - - -

As expected, no negative roots. So we are sure there is one positive real root, and perhaps three of them.

So amongst the possibilities are 1, 3, 9, 27, and 1/2, 3/2, 9/2, 27/2.

αβγ=27/2=13.5

According to Descartes, all the real roots are positive. Considering our possible rational roots from above, it seems hard to find a possible combination. We know for sure that at least one root is real and positive, so perhaps the other roots are complex numbers of the form a+bi and a-bi, where a is negative.

In this case, our real root must be greater than +2. The first number we have greater than 2 is 3, which we can try in Horner's scheme, or otherwise.

2 | -4 | 3 | -27 | |

3 | 0 | 6 | 6 | 27 |

2 | 2 | 9 | 0 |

2x

Because b

The roots are:

x1= – 0.5 + 2.0615528i

x2= – 0.5 – 2.0615528i

Or x= -0.5 ± √(17/4)i

+ + + +

There aren't any positive real roots.

As every cubic has 3 roots, then, we know immediately that this one has 3 negative real roots or one.

Just for the record, the signs of the coefficients for f(-x) are:

- + - +

Showing 3 sign changes and indicating 3 or one negative real roots.

αβγ=-12/84=-1/7

All the real roots are negative, and one root is definitely real and negative.

+ - - -

There is definitely one positive real root. This means that because a cubic has 3 roots, there are 2 or 0 real negative roots. The root near zero is therefore a negative number. From the graph, we see the function has 3 real roots.

The possible denominators, q, are:

1.1.9

1.3.3

So at least one root would be a whole number.

and the possible numerators, p are:

1.17.25

1.1.425

5,5,17

We can quickly (or more quickly) eliminate q=±1, with 1 as a denominator, leaving ±5, ±17 and ±25 to check. However, if ±5 is a numerator, then the other roots are ±5 and ±17. Otherwise, if ±25 is a numerator, then the other is ±17. (We need to use up all the factors), so ±17 is worth checking first.

αβγ=425/9=47+2/9 [2]

One root is positive and the other 2 are negative. It seems that, also looking at the Rational Root Theorem and the graph, one root is 25, and one is -17 . The other root seems to be -1/9 because it needs to be small, and 25-17=8, and to get the sum of all the roots [1], subtracting 1/9 gives us the right answer.

Also, the product of these roots is: 25(-17)(-1/9)=425/9 which is the same as [2].

We try Horner's Scheme for -1/9, and because it works, we also try 25

9 | -71 | -3833 | -425 | |

-1/9 | 0 | -1 | 8 | 425 |

9 | -72 | -3825 | 0 | |

25 | 0 | 225 | 3825 | |

9 | 153 | 0 |

The last line, 9,153 is 9x+153=0, so x=-17

The roots are -17, -1/9 and 25

The graph indicates that there are three real roots near -1.5, 0 and 1.

The signs of the coefficients are:

+ + - +

There are two positive real roots, or none. The root near zero is therefore a positive real root, as the other positive real root is near 1.

The signs for f(-x), used to find the negative real roots (which we already know!) are:

- + + +

As expected, there is one negative real root.

α+β+γ= -59/100 [1]

αβγ=-2/100 [2]

Zooming in to between -1.5399949 and -1.5399948, suggests the root is very close to -1.5399948

The approximate root by Cardano's formula is –1.5399948019

The roots of the equation 100x³+59x²–145x+2 are:

x1= 0.9361215427

x2=–1.5399948019

x3= 0.0138732592

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