A Nice Cubic Equation

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[seanfraser]

By inspection or using the rational root theorem, we find that x=-2 is a solution.

We can now factor this polynomial:
(x+2)(x^2-2x+9)

Use the quadratic formula to find the other roots:

x = (2 +- sqrt(2^2 -4(9)))/2

x = 1 +- 2sqrt(2)i

So the solutions are x=-2, x=1+2sqrt(2)i, and x=1-2sqrt(2)i

[DonRedmond-jkhj]

Descartes Rule of Signd, Rational Root Theprem and Synthetic Division

[sollevi]

-2 by inspection after 5 seconds then solve the remain quadratic

[Qermaq]

Rational Root Theorem, tried a few factors, found one that worked, and bang.

[NadiehFan]

Nice, this is what your previous misguided video about solving the cubic x³ + 2x² + 9 = 0 where you failed to find its rational root x = −3 should have been like. Glad to see that you removed that video.

Note, however, that your second 'method', which is about using the method of undetermined coefficients to try to factor a cubic as a product of a linear and a quadratic polynomial, is bound to fail. This will always result in a cubic equation which is either identical to or equivalent with the cubic you are trying to solve in the first place, i.e. this will only get you back to square one (or rather, cube one). At 8:22 you have arrived at

18/a − a² = 5

but this equivalent with

a³ + 5a − 18 = 0

and substituting a = −d this gives

d³ + 5d + 18 = 0

which is of course _exactly_ the equation x³ + 5x + 18 = 0 you started with, only with a different name for the variable. So, don't recommend the method of undetermined coefficients for factoring cubics into a linear and a quadratic polynomial as this will only get you back where you started, meaning it is a complete waste of time.

What you should have done here is apply the rational root theorem right away. Since x³ + 5x + 18 = 0 cannot have any positive real roots we only need to test −1, −2, −3, −6, −9, −18 and then we quickly find that x = −2 is a solution.

Note that x³ + 5x + 18 is _strictly increasing_ on the real number line, so the equation x³ + 5x + 18 = 0 has _exactly one_ real root. This means that its other two roots are conjugate complex roots and that we don't need to try other potential rational solutions once we have found that x = −2 is a solution.

To factor the equation

(1) x³ + 5x + 18 = 0

once we have found that x = −2 is a solution because

(2) (−2)³ + 5·(−2) + 18 = 0

we can simply subtract (2) from (1) because both sides of (2) are equal, which gives

x³ + 2³ + 5x + 5·2 = 0

where we can take out a factor x + 2 which is common to x³ + 2³ and 5x + 5·2 to get

(x + 2)(x² − 2x + 2² + 5) = 0

which is

(x + 2)(x² − 2x + 9) = 0

[devondevon]

my method
Answer 2 and 1 + 2.82843i and 1 - 2.82843i (imaginary numbers)
x^3 + 5x + 18 = 0
x^3 + 2^3 + 5x +10 = 0 (as 2^3 + 10 =18)
but x^3 + 2^3 can be expressed as (x+2)( x^2 + 2^2 - 2x)
Hence, (x+2)(x^2 + 2^2 -2x) + 5x + 10 =0
(x+2)(x^2 + 2^2-2x) + 5(x+2) = 0 ( 5x +10 = 5(x+2)
x^2 + 4-2x+ 5 (x+2)=0 as p(n) + p (y) = p(n+y)
x^2 + 9- 2x (x+2) =0
Hence x^2 -2x + 9 = 0
solving for x = 1 + 2.82843i imaginary answer
and x =1 - 2.82843i imaginary number answer
but x + 2 =0
Hence, x also = -2 Answer real number

[alnitaka]

If you use the cubic formula you get If you look for roots among the divisors of 18, you get that -2 is a root. So is a complicated way of saying -2. The temperature was degrees Celsius this morning outside my house. If you stick omegas in as factors of the two terms, you get the other two roots, 1 + or - 2i*sqrt(2).

[goldfing]

Why not just test the divisors of 18? From Descartes' rule of signs, it is clear that a real solution must be negative, and x = -2 is easily been found, since (-2)^3 + 5*(-2) + 18 = -8 - 10 + 18 = 0.
Then do a polymial division to get x^2 - 2x + 9 = 0 as the remaining quadratic to solve.

[braydentaylor]

Request:
x^18-(x^6-x^2)^3-(x^5-1)^3=1, find all real and complex solutions

[kareolaussen]

Strategy to solve quadratic equations:
1. Search for rational (often integer) roots, using the rational root theorem.
If point 1 does not work:
2. Complete the square (I consider it brain dead to use the quadratic formula).

Strategy to solve cubic equations:
1. Search for rational (often integer) roots, using the rational root theorem.
If point 1 does not work,
2a. Reduce to case x^3 + bx + c = 0, if necessary.
2b. Write x=u+k/u, and choose k to obtain a quadratic equation in u^3. It is not necessary to remember that k=-b/3, but it is useful to note that the two roots of the quadratic equation is related by
u_2^3 = (k/u_1)^3,
which means that we may write the solution as
x=u_1+u_2.

Strategy to solve quartic equations:
1. Search for rational (often integer) roots, using the rational root theorem.
If point 1 does not work,
2a. Reduce to form x^4 = -bx^2 -cx -d, if necessary.
2b. Rewrite as (x^2 + k)^2 = (2k-b)x^2 -cx + k^2-d, and choose k such that the right hand side is a perfect square in x. This requires the discriminant to vanish, which is a reduced cubic equation for k. With this rewriting the quartic equation may be written as (x^2+k)^2 - (2k-b)(x-q)^2 = 0, which can be rewritten as the product of two quadratic factors.

[rakenzarnsworld]

x = -2
(x+2)(x^2-2x+9)=0
x = 1+-{2^(3/2)}i

[bkkboy-cmeb]

x³+5x+18=0
(x+2)(x²-2x+9)=0
∴x=-2, x=1±2√2i

[prollysine]

x^3+/-x^2+5x+18=0, (x+2)(x^2-2x+9)=0, x^2-2x+9=0, x=(2+/-V(4-36))/2, x=(2+/-i*V32)/2, x= 1+/-i*V8,
1 2 solu, x= -2, 1+i*2*V2, 1-i*2*V2,
-2 -4
9 18

[kareolaussen]

Bypassing rational root search, let me try to find the real solution of this equation by the general method:
Write x = u+k/u, with k=-5/3 to get the equation
u^3 + k^3/u^3 + 18 =0, or
(u^3)^2 +18 u^3 + k^3 = 0, or
(u^3 + 9)^2 = 81-k^3 =2312/27=(34/9)^2 * 6.
I.e
u^3 = (34√6-81)/9 =((2√6-3)/3)^3, or
u^3 =-(34√6+81)/9=-((2√6+3)/3)^3.
Hence we find real roots
u_1 = (2/3)√6-1, u_2 = -(2/3)√6-1.
Note that u_2 = k/u_1 =-5/(3u_1), which is a general property, leading to
x = u_1 + u_2 = -2.
To find the simplified expressions for u (i.e without the nested roots) is much harder than searching for rational roots of the original equation!

[Blaqjaqshellaq]

The complex solutions are 1 plus or minus i*2*2^1/2.