How do you factor the trinomial $x^2-11xy+60y^2$ ?

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Answer 1 · 2016-01-04T10:46:14

$x^2-11xy+60y^2$

$= 1/4(2x-(11+sqrt(119) i)y)(2x-(11-sqrt(119) i)y)$

$x^2-11xy+60y^2$ is in the form $ax^2+bxy+cy^2$ with $a=1$ , $b=-11$ and $c=60$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-11)^2-(4*1*60) = 121-240 = -119$

Since $Delta < 0$ this trinomial has no linear factors with Real coefficients.

It does have factors of the form $(x-t_1y)(x-t_2y)$ where $t_1$ and $t_2$ are the (Complex) roots of $t^2-11t+60 = 0$ given by the quadratic formula:

$t = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-b+-sqrt(Delta))/(2a)$

$=(11+-sqrt(-119))/2$

$=11/2+-sqrt(119)/2 i$

Hence:

$x^2-11xy+60y^2$

$= (x-(11/2+sqrt(119)/2 i)y)(x-(11/2-sqrt(119)/2 i)y)$

$= 1/4(2x-(11+sqrt(119) i)y)(2x-(11-sqrt(119) i)y)$

How do you factor the trinomial x^2-11xy+60y^2 x^2-11xy+60y^2 ?