Solving a Spectacular Factorial Equation | Algebra

Solving a Spectacular Factorial Equation | Algebra

In this captivating video, we delve into the intriguing realm of factorial equations, unlocking their secrets and cracking the code to solve a truly spectacular factorial equation. Join us on a mathematical adventure as we unravel the complexities and reveal the step-by-step process to conquer this captivating challenge. With insightful explanations, factorials, cubic equation, remainder theorem and expert tips, you'll gain the knowledge and skills to confidently solve factorial equations and witness the remarkable power of factorials in action. Get ready to crack the factorial code and unravel the mysteries of this spectacular equation!

Topics Covered:
Algebra
Factorials
Solving Factorial equations
Cubic Equation
Remainder theorem
Quadratic equation
Quadratic formula
Algebra Challenge

8 Key moments of this video:
0:00 Introduction
0:36 Factorials
1:26 Cubic equation
3:45 Finding one factor
4:30 Remainder theorem
6:12 Quadratic equation
6:24 Quadratic formula (Discriminant)
7:38 Verification

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Thanks for Watching!!
@infyGyan

[TypoKnig]

Good demonstration of factoring and solving a cubic! A shortcut I took was to recognize that the left had side was the product of 3 consecutive integers. Therefore the right hand side must also be the product of 3 consecutive integers. The prime factors of 210 are 2, 3, 5, and 7. Notice 2*3 = 6, so 210 = 5*6*7. The smallest term is x+3 and = 5, so x = 2.

[ChavoMysterio]

5:00 I would use polynomial or synthetic division to get the other factor.

[walterwen]

Solving a Spectacular Factorial Equation: (x + 5)!/(x + 2)! = 210; x = ?
(x + 5)!/(x + 2)! = 210; When x is a positive integer, x ≥ 0, otherwise x is a complex number
First method:
[(x + 5)(x + 4)(x + 3)(x + 2)!]/(x + 2)! = (x + 5)(x + 4)(x + 3) = 210 = (7)(30) = (7)(6)(5)
(x + 5)(x + 4)(x + 3) = (2 + 5)(2 + 4)(2 + 3); x = 2
Missing two complex roots
Second method:
Let: y = x + 5, x + 4 = y – 1, x + 3 = y – 2
(x + 5)(x + 4)(x + 3) = y(y – 1)(y – 2) = 210, y^3 – 3y^2 + 2y – 210 = 0
(y^3 – 7^3) – (3y^2 – 2y – 133) = (y – 7)(y^2 + 7y + 49) – (y – 7)(3y + 19) = 0
(y – 7)(y^2 + 7y + 49 – 3y – 19) = (y – 7)(y^2 + 4y + 30) = 0, y – 7 = 0 or y^2 + 4y + 30 = 0
y = 7, x = 2 or y^2 + 4y + 4 = (y + 2)^2 = – 26; y = – 2 ± i√26, x = – 7 ± i√26; Complex roots
Answer check:
x = 2, (x + 5)!/(x + 2)! = (x + 5)(x + 4)(x + 3) = 210; Confirmed in First method
x = – 7 ± i√26; y^2 + 4y + 30 = 0, y^2 = – 4y – 30, y = x + 5
(x + 5)(x + 4)(x + 3) = y(y – 1)(y – 2) = y(y^2 – 3y + 2) = y(– 4y – 30 – 3y + 2)
= (– 7y)(y + 4) = (– 7)(y^2 + 4y) = (– 7)(– 30) = 210; Confirmed
Final answer:
x = 2, x = – 7 + i√26 or x = – 7 – i√26