This is the answer in the book but I don't know how they got from

x^2-5root(2)X+12=0

to

(x-3root(2))(x-2root(2))

Help please?

Thank you!

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- February 6th 2009, 06:07 PMpositiveionHow to solve algebraic surds by factorising?
This is the answer in the book but I don't know how they got from

x^2-5root(2)X+12=0

to

(x-3root(2))(x-2root(2))

Help please?

Thank you! - February 6th 2009, 07:02 PMelliotyang
well, you can use the quadratic formula to find the roots of the equation, which the roots are

[-b+sqrt(b^2-4ac)]/2a and [-b-sqrt(b^2-4ac)]/2a

where a=1, b=-5root(2), c=12

after you get the roots, rewrite your equation in the form of

(x-root1)(x-root2)=0 - February 7th 2009, 06:16 AMHallsofIvy
First note that . That means that in order to factor a polynomial of the form you look for numbers and such that and . In this particular example and so you need to solve and . Since that product does NOT have a in it, it is clear that both and must have a factor of .

So, now, we are thinking and for some numbers m and n. and so we now have m+ n= 5 and mn= 6. -2 and -3 satisfy those equations so we can take and and have