A Nice Polynomial Equation | Cubic Formula

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[nguyenphungdunganh]

for a polynomial, if the sum of coefficients equal zero, then one of the roots is one, if the sum of even power coefficients minus the sum of odd power coefficients, the one of the roots is -1. here we have x^3+x^2-2=0 which means the sum of coefficients is 0 so x=1 is a root. then polynomial division will get you the complex roots

[FrogworfKnight]

Done before watching the video. ` added for spacing during synthetic division since youtube doesn't like extra spaces.
x^3+x^2=2

x^3+x^2-2=0

1| 1 1 0 -2
` | `1 2 2

` 1 2 2| 0
x^2+2x+2=0
x=(-2+-sqrt(4-8))/2
x=(-2+-2i)/2
x=-1+-i

x=1, -1+i, -1-i

[kicorse]

In case anyone is wondering where the complex roots where skipped over in the first method, it was in getting from c_1 and c_2 to a and b. Here, there are three solutions to a^3 = c_1, for example.

[charlesmrader]

I like to start a problem like this with a guess. Guess that the solutions for x are integers. Immediately we see x=1 is a solution. Then we do synthetic division by (x-1) and we get a quadratic which we can solve with the quadratic formula.

[qwang]

By the rational root theorem, possible rational solutions are 1, -1, 2, -2. Only x = 1 is the solution. By synthetic division, one can easily find x^2 + 2x + 2.
So x^3 + x^2 - 2 = (x-1)(x^2 + 2x + 2) = 0. x^2 + 2x + 2 = has 2 complex conjugate solutions.

[hamidkh]

Thank you proffessor.
by the way in your first method, in ax^3+bx^2+cx+d=0 when c = 0 , then we can use x = 1/y . it's easier

[SteveOmnipotent]

For such equations of higher degree, where we know that the positive solution will be 1. We can multiply the integer(number on RHS) by the variable.
For example:
x^3 + x^2 + x = 3
or, x(x^2 + x + 1) = 3(x)
or, x^2 + x = 2
or, x(x + 1) = 2(x)
or, x = 1

[alipourzand]

In the first method, the sustitution constant is actually the inflection point of the cubic function.
f(x) = x^3 + x^2 - 2
f''(x) = 6x + 2
f''(x) = 0 -> x = -1/3

[sumit]

2nd method is the easy one and that's what I used, this was too easy

[rickdesper]

OK, I'm not going to go down the rabbit hole here. But the consensus seems to be: if you see x^3 + x^2 = 2, the obvious path to take is to note that x=1 is a root. Factor out (x-1) from x^3 + x^2 = 2 = 0 = (x-1)p(x), and use the quadratic formula to get the two roots of p(x). Surely that's an easier path to take than using a variable substitution and the "formula for cubics". There is no shame in using an obvious root, especially in a case where the rational root theorem says that only four rational numbers could be roots: +/-1 and +/-2.

[zawatsky]

Не поленился досмотреть до конца только затем, чтобы убедиться, что в конце действительно будет единица. Автор отлично справляется с ролью Капитана Очевидности. )

[RexxSchneider]

The rational root theorem indicates possible integer roots at { ±1, ±2), and once you spot that the sum of the coefficients is 0, you know that x=1 is a solution. So divide (x^3+x^2-2) by (x-1) and you get x^2+2x+2. The quadratic formula the yields the other two solutions: (-2 ± √(4 - 8) ) / 2 = -1 ± i.
You should demonstrate polynomial division once in a while, as an alternative to your "round the houses" factorisation method. It's really no worse than conventional long division.

[samsunga-rrfw]

Excellent! So welll explained! More please!😂❤

[hamster]

If X>1, then X³+X²>2
If X<1, then X³+X²<2
Do we need something else?
To prove it, it is quite possible to use the signs of inequality too > (more) and < (less).
For example, X²-X=1 excludes 0 and 1.

[ralphaad]

x^3 + x^2 = 2
(x^2)(x + 1) = 2
Factors of 2 are 1 and 2
1^2 = 1 and 1 + 1 = 2
1 × 2 = 2
x=1

[mcwulf]

Obvs x-1 is a factor. Take that out and solve the quadratic that remains.

[goldfing]

Hmm, why do you title "cubic formula" when there is an easy to find an integer solution x = 1 and you dont need the cubic formula?
1^3 + 1^2 = 1 + 1 = 2, so x1 = 1 is a solution and we can divide
(x^3 + x^2 - 2):(x - 1) = x^2 + 2x + 2
And solve
x^2 + 2x + 2 = 0
by subtracting one:
x^2 + 2x + 1 = -1
And using the binomic formula:
(x + 1)^2 = -1
x + 1 = +- sqrt(-1)= +- i
x2, 3 = -1 +- i
I cannot think of any easier way to solve this equation.

[mdshamimkhan]

X³+X²=2, let X² be Y
Therefore, Y+1=2
Y=1
Therefore,
X²=1
So X=±1, but -1 is omitted as (-1)³= -1,
So X=1....

[christiansimon]

x=1 is immediatly seen. So divide by x-1 and easy

[ynfnvrm]

Obviously 1 is a root. So you can by Vieta's formula other two roots