
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 16th, 2019, 09:07 AM  #1 
Newbie Joined: Jan 2018 From: Belgium Posts: 8 Thanks: 0  Factor polynomial
Hello, When making some exercices to improve my general understanding of basic mathematics, I stumbled upon following problem: to factorise X^2 + ((sqrt 2)+(sqrt 3)) x + (sqrt 6) I do not recognise the form (a+b)^2, so I tried going for the discriminant, which would be ((sqrt 2)+(sqrt 3))^2  4*(sqrt 6) = 2 + 3 + 2((sqrt 2)+(sqrt 3))  4 (sqrt 6) = 5  2(sqrt6) I thus there are two roots to this polynomial, but I do not know how to calculate them. When trying to use the formula (b + (sqrt D))/2a, the numbers (especially the square roots) get to hard for me to work with. The book lists the solutions (x+ (sqrt 2)) * (x+ (sqrt 3)), but not how to get there. Could anyone give me a hint here please? 
February 16th, 2019, 09:13 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,342 Thanks: 984 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
February 16th, 2019, 09:26 AM  #3 
Newbie Joined: Jan 2018 From: Belgium Posts: 8 Thanks: 0 
Oh, wow, I didn't know that equation. Thank you!

February 16th, 2019, 10:21 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,342 Thanks: 984 Math Focus: Wibbly wobbly timeywimey stuff.  
February 16th, 2019, 01:21 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,342 Thanks: 984 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle x^2 + ( \sqrt{3} + \sqrt{2} )x + \sqrt{6} = 0$ $\displaystyle x = \dfrac{(\sqrt{3} + \sqrt{2} ) \pm \sqrt{ ( \sqrt{3} + \sqrt{2} )^2  4 (1) \sqrt{6} }}{2 (1)}$ $\displaystyle x = \dfrac{  ( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2 + 2\sqrt{6}  4 \sqrt{6} } }{2}$ $\displaystyle x = \dfrac{ ( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2  2 \sqrt{6} } }{2}$ $\displaystyle x = \dfrac{ (\sqrt{3} + \sqrt{2} ) \pm \sqrt{(\sqrt{3}  \sqrt{2})^2}}{2}$ $\displaystyle x = \dfrac{ (\sqrt{3} + \sqrt{2} ) \pm (\sqrt{3}  \sqrt{2} )}{2}$ I leave it to you to finish this. Dan  

Tags 
factor, polynomial 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How to factor this polynomial  shiseonji  Algebra  3  August 4th, 2013 01:12 PM 
Factor this second degree polynomial  mrmorris  Algebra  5  August 30th, 2011 12:36 PM 
Why can't 3/2 be a factor of this polynomial?  figgy  Algebra  5  June 12th, 2011 07:31 AM 
factor polynomial  noname  Abstract Algebra  2  November 17th, 2006 08:39 AM 