Solving a Cubic Equation in Two Ways (x^3-x^2-4=0)

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#ChallengingMathProblems #PolynomialEquations

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[MathTutor]

This is a good one. You can solve this problem in four ways: Two using grouping, one using factor theorem and another using Cardano's formula (typical for cubic).

[VSN]

I simply realised x=2 is a solution then by Long division and discriminent determine the quadratic to be complex. I really like both your methods though! Although not too useful in this question but in equations with non-integer roots, I can see this as a fairly useful technique!

[v._.v]

I have another way:
f(2)=0, therefore x-2 is a factor. (Since the constant 4’s factors are 1, 2, 4, so I subbed all 3 numbers into x)
By comparing coefficient, f(x)=(x-2)(x^2+x+2)
Solve f(x)=0. Done.

[carloshuertas]

Another great explanation, SyberMath! Before starting watching this video, I actually knew that x=2 is one of the solutions. Thanks a lot!

[pixcalibur]

Inspection can give x=2 as a solution.
Factoring an (x-2) piece out gives x^3-x^2-4=0 -> (x-2)(x^2+x+2)=0
Solutions:
x=2
x=[-1 + i*root(7)]/2
x=[-1 - i*root(7)]/2

[WahranRai]

Vieta formula for P(x) = ax^3 + bx^2 + cx + d with x, y, z the 3 roots :
the given polynomial ( x^3 -x^2 - 4 = 0) has : a=1; b=-1; c=0; d=-4
x + y+z = -b/a = 1
xy + xz + yz = c/a = 0
xyz = -d/a = 4
x = 2 is an obvious solution of given polynomial. Then equations 1 and 3 implies :
y+ z = -1 and yz = 2
We have to find 2 numbers knowing their sum S = -1 and product P =2 :
u^2 - Su + P = 0 = u^2 + u + 2 etc...

[GUTY]

A ojo se ve que una solución es 2 . Con Ruffini y la resolvente podes encontrar las otras soluciones.

[pfeqpsb]

まともに解くのは難しい。
x=2が解だから（x-2）で全体が割れるのだが
この動画の解法は巧妙。

[scottleung]

I wasn't sure how to factor it, but I did get x=2 as the only real solution by plugging it in.

[holyshit]

x=2 by guessing
x^3-x^2-4 =0, by grouping
x^3-8-x^2+4=0
(x-2)(x^2+2x+4)-(x-2)(x+2)
(x-2)(x^2+x+2)=0
x^3-x^2-4=0
Recall binomial expansion (it will be also useful later)
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
so if we take a=x then we have to choose b=-1/3
(x - 1/3)^3=x^3 - x^2 + 1/3 x -1/27
Now after expansion we have got 1/3x but in original equation we have 0x so
(x - 1/3)^3 - 1/3(x-1/3) = (x^3 - x^2 + 1/3 x -1/27)-(1/3x-1/9)
(x - 1/3)^3 - 1/3(x-1/3) =x^3 - x^2 +2/27
(x - 1/3)^3 - 1/3(x-1/3)-110/27=0
y^3-1/3y-110/27=0
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
(a+b)^3 = a^3+3ab(a+b)+b^3
(a+b)^3 - 3ab(a+b) - (a^3+b^3) = 0

ab=1/9
a^3+b^3=110/27

a^3b^3=1/729
a^3+b^3=110/27

t^2-110/27t+1/729=0
(t-55/27)^2-3024/729=0
56*54=4*14*6*9=144*21


x - 1/3 =
x =
x =
x =
x =
x = 1/3 * 6
x = 2

[ahmadmazbouh]

0:45 wasn't Ferrari a car? i've never heard about this tbh, i'll search about it asap, thx

[stvp]

Always love hearing you say “two methods”🧡🧡

[JohnRandomness]

I have to admit that I zeroed in on "two" as a solution -- 2³ - 2² - 4 = 8 - 4 - 4 = 0 -- way too fast. I followed the brute-force cubic method, but then had no idea how to simplify the solution except with a calculator.

I did notice another thing: if I subtract and add 4x, I get 0 = x³ - 4x - x² + 4x - 4 = x(x-2)(x+2) - (x-2)² = (x-2)[x(x+2) - (x-2)]. Then the bracketed factor is x² + x + 2, which has no real root.

So x = 2 is the only real solution. (I got that from the brute-force method as well.)

[chaosredefined]

If I encounter this in exam conditions, then I'm probably just going to use the rational root theorem... The other ones look nice and all, but I just need to find one of the roots. Once I have one, I can factor it out and find the other two.

[gtziavelis]

Desmos has a reverse contrast mode which would fit in great with SyberMath dark background!

[ahmadmazbouh]

i got the x=2 in the first (π×10^-π) seconds so i factored out x-2 using factoring by grouping then i did the rest using the formula

[-basicmaths]

x^3-x^2=4
x^2(x-1)=4*1
x^2=4 and x-1=1
x=2

[cenkalmastal]

This problem;
x^3-x^2-4=0; => x^3-x^2=4; => x^2*(x-1)=4*1; => x^2=4; (x-1)=1;
that's solved x1=+2; x2=-2; x3=+2; good days!

[rakenzarnsworld]

Find one real root and leftover two roots could be solved instantly.

[tonyhaddad]

First method is awesome, this kind of approach satisfy me a lot, that s what i always expect from math 😍