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 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
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 Originally Posted by Atrend 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?
$\displaystyle \sqrt{6} = \sqrt{2} \cdot \sqrt{3}$ so the quadratic is of the form $\displaystyle x^2 + (a + b)x + ab$ which factors as $\displaystyle (x + a)(x + b)$.

-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
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 Originally Posted by Atrend Oh, wow, I didn't know that equation. Thank you!
You probably do. For example, factor $\displaystyle x^2 + 3x + 2$. It's the same, just with radicals.

-Dan February 16th, 2019, 01:21 PM   #5
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Quote:
 Originally Posted by Atrend 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?
Just to cover all bases:
$\displaystyle x^2 + ( \sqrt{3} + \sqrt{2} )x + \sqrt{6} = 0$

$\displaystyle x = \dfrac{-(\sqrt{3} + \sqrt{2} ) \pm \sqrt{ ( \sqrt{3} + \sqrt{2} )^2 - 4 (1) \sqrt{6} }}{2 (1)}$

$\displaystyle x = \dfrac{ - ( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2 + 2\sqrt{6} - 4 \sqrt{6} } }{2}$

$\displaystyle x = \dfrac{ -( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2 - 2 \sqrt{6} } }{2}$

$\displaystyle x = \dfrac{ -(\sqrt{3} + \sqrt{2} ) \pm \sqrt{(\sqrt{3} - \sqrt{2})^2}}{2}$

$\displaystyle x = \dfrac{ -(\sqrt{3} + \sqrt{2} ) \pm (\sqrt{3} - \sqrt{2} )}{2}$

I leave it to you to finish this.

-Dan Tags factor, polynomial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shiseonji Algebra 3 August 4th, 2013 01:12 PM mrmorris Algebra 5 August 30th, 2011 12:36 PM figgy Algebra 5 June 12th, 2011 07:31 AM noname Abstract Algebra 2 November 17th, 2006 08:39 AM

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