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 February 16th, 2019, 08:07 AM #1 Newbie   Joined: Jan 2018 From: Belgium Posts: 7 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, 08:13 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, 08:26 AM #3 Newbie   Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0 Oh, wow, I didn't know that equation. Thank you!
February 16th, 2019, 09:21 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, 12:21 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

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