Solving a Quartic Equation

You should not take calculus if you can't solve this. This video utilizes the most common strategies adopted in solving calculus problems. I used substitution, factoring, theorems, synthetic division and the quadratic formula

[rimantasri]

It's great to see how such many mathematical theorems or manipulations appear in one simple-looking problem! You explained it flawlessly!

[elio]

it is better to use "1 1" at the top of the Pascal triangle. This way. "1 2. 1" follow the rule

[emanuelborja]

I substituted y just as you did. Then I subtracted 2 from each side, and grouped it as ((y-1)^2-1) + y^3 + ((y+1)^2-1). This allowed me to factor the two groups as a difference of squares. Y then factored out nicely as a common factor, revealing y=0 as a solution. A bit of further work also allowed (y+2) to factor out nicely, revealing y=-2 as a solution. The remaining quadratic was solved just as you did. This method avoided the need for polynomial division and distribution of the quartic term.

[nstarling]

You have a forever subscriber from this video. As a math teacher my self this was flawless.

[TheRenaSystem]

I've been binging your channel since discovering it, and just wanted to say I love your style of presentation and how you teach, 10/10.

You have perfect handwriting, great explanatory skills, you speak clearly, and you have a really soothing voice to boot lol.

[laman]

Great. Never seen this solution before. Very interesting. We have watched a number of clips by this dude. His patience and systematic approach are excellent. Born to be a teacher.

[gregmackinnon]

Great teaching. So many techniques in one problem. Brilliant!

[petejackson]

I used y=x + 3 instead and it was a lot easier using a difference of 2 squares as part of the factorisation .

[nullplan]

Tried this before watching:
(x+1)² + (x+2)³ + (x+3)⁴ = 2

All coefficients will be integers (obviously). Thus we get to use the rational roots theorem.
Lead coefficient will be 1.
Constant part will be 1² + 2³ + 3⁴ - 2 = 88.

So, only possible rational roots are the divisors of 88 (positive and negative, of course).

88 = 2³ * 11, so divisors of 88 are 1, 2, 4, 8, 11, 22, 44, 88

Attempt at x = 1: (x+1)² = 4, adding further positive numbers will not decrease the value. No, we need negative numbers.

Attempt at x = -1: (0)² + (1)³ + (2)⁴ = 1 + 16 = 17 ≠ 2.
Attempt at x = -2: (-1)² + (0)³ + (1)⁴ = 2. Winner!
Attempt at x = -4: (-3)² + (-2)³ + (-1)⁴ = 9 - 8 + 1 = 2. Winner!
Attempt at x = -8: (-7)² + (-6)³ + (-5)⁴ = 49 - 216 + 15625 ≠ 2.

The quartic term has far outpaced the cubic one at this point. Going lower will not help.

So, it is time to pay the piper and face the music:

(x+1)² + (x+2)³ + (x+3)⁴ = 2
x² + 2x + 1 + x³ + 6x² + 12x + 8 + x⁴ + 12x³ + 54x² + 108x + 81 = 2
x⁴ + 13x³ + 61x² + 122x + 88 = 0

(x⁴ + 13x³ + 61x² + 122x + 88) : (x + 2) = x³ + 11x² + 39x + 44
(x³ + 11x² + 39x + 44) : (x + 4) = x² + 7x + 11


We can solve the
x = -7/2 ± √(49/4 - 11) = -7/2 ± √(49/4 - 44/4) = -7/2 ± √5/2


Thus the solutions are:
x₁ = -2
x₂ = -4
x₃ = (-7-√5)/2
x₄ = (-7+√5)/2

[chao.m]

This is way cool. Never seen the rational root theorem and remainder theorem explained and applied in such simple and easy to follow manner. Also that long division with an addition rather than subtraction is excellent too. Goes to show the difference between teaching and effective teaching: solve complex problems without breaking a sweat

[robertlezama]

Excellent delivery ... love his manner of teaching, he makes math seem fun and not scary. More importantly, he applies techniques using the formal theorem names, so if you need to brush up you can go find the theorems and study them outside of solving actual problems. Really well done. Thanks!

[DaveyJonesLocka]

I like this. You could solve it by brute force, but taking advantage of specific characteristics of a problem to unravel a more elegant solution is just prettier.

I do a lot of driving, and often try to solve problems like this one mentally. This was a fun one to do.

[vikasseth]

You are the coolest maths teacher I have seen. Super

[KG_]

Consider this approach..
(to avoid expanding 4th power)
let x+3 = y
Equation becomes (y-2)^2 + (y-1)^3 + y^4 = 2
after expanding using identities equation becomes
y^4 + y^3 - 2y^2 - y + 1 = 0
Rearranging the terms makes it easy to factorize
(y^4 - 2y^2 + 1) + (y^3 - y) = 0
(y^2 - 1)^2 + y(y^2 - 1) = 0
(y^2 - 1) {y^2 - 1 + y} = 0
problem is 90% done 😃

[juanrobles]

I am amaze on how you broke down by explaining key concepts and theorems to justify your answer. Great explanation sir.

[sankararaopulla]

Nice selection of sums and a wonderful explonation

[nicolascamargo]

Otra alternativa:
(x+1)²+(x+2)³+(x+3)⁴=2
Suma de potencias de tres números consecutivos luego 2=1+1=(-1)²+0³+1⁴, así x+1 debe ser -1 en está solución de donde x=-2 así (x+2) es factor y se puede hacer lo siguiente:
(x+1)²+(x+2)³+(x+3)⁴=2
(x+2-1)²+(x+2)³+(x+2+1)⁴=2



Los coeficientes del polinomio en x+2 son 1, 5, 7, 2, posibles raíces racionales para x+2 son: ±1 y ±2.
Como 1+7≠5+2 queda descartado x+2=-1
Como 1+5+7+2=15≠0 queda descartado x+2=1.
Como queda descartado x+2=2.
Como (-2)³+5(-2)²+7(-2)+2=-8+20-14+2=22-22=0, x+2=-2 funciona así x=-4 es solución y por lo tanto (x+4) es factor así se puede hacer lo siguiente:
(x+1)²+(x+2)³+(x+3)⁴=2




(x+2)[(x+4)³-(x+4)²-(x+4)]=0
(x+2)(x+4)[(x+4)²-(x+4)-1]=0



Así:
(x+1)²+(x+2)³+(x+3)⁴=2
Tiene como soluciones:
x_1=-2, x_2=-4, x_3=-(7/2)-((√5)/2) y x_4=-(7/2)+((√5)/2).

[SanePerson]

Nice, clear exposition (and extremely nice board work!). I have few personal comments from the perspective of someone who took algebra and precalculus 50+ years ago and used it (along with trigonometry, calculus, linear algebra, differential equations, and various other upper division math) in my career as a theoretical chemist. I disagree with the opening note, "You should not take calculus if you can't solve this.") This is nonsense, BUT it is correct to say, "If you've forgotten how to do this, be prepared to relearn it when you take calculus" Almost all high school math will be needed for more advanced math, but you learn what is most important by the necessity of USING it. Personally, I always made fewer errors doing long division into polynomials than I made trying to remember exactly how synthetic division works. Pre-calc teachers use synthetic division all the time, but people who apply math, only come across these kinds of problems occasionally - tried-and-true long division I always remembered - synthetic division got hazy. By the same token, students will absolutely need trigonometry in doing calculus, but all those identities? Few people remember them. Just be prepared to relearn the most important ones when you're learning calculus.

[ndotl]

Subscribed because you explain why each step was taken, which frees the learner from the rote memorization form of education.

[pierreneau]

Thanks to share such equation. Other way: finding trivial solutions. It's necessary to check with low value of power 4. X=-2 and x=-4 can be easily found. After that, it's necessary to develop x^4+13x^3+61x^2+122x+88=0 and factorize by (x+2) and (x+4) to obtain (x+2)(x+4)(x^2+7x+11)=0 The last 2 solutions are (-7+-sqrt(5))/2.