How do you factor $6x^2y + 36xy - 66y$ ?

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Answer 1 · 2015-05-19T11:50:01

All the terms are multiples of $6y$ , so we find:

$6x^2y+36xy-66y = 6y(x^2+6x-11)$

$x^2+6x-11$ is of the form $ax^2+bx+c$ with $a=1$ , $b=6$ and $c=-11$ , which has discriminant given by the formula:

$Delta = b^2-4ac = 6^2-(4xx1xx-11) = 36+44 = 80$

This is positive, but not a perfect square. So the zeros of $x^2+6x-11$ are not rational and its linear factors will have irrational coefficients.

The roots of $x^2+6x-11 = 0$ are given by the formula:

$x=(-b+-sqrt(Delta))/(2a) = (-6+-sqrt(80))/2$

$=(-6+-4sqrt(5))/2$

$=-3+-2sqrt5$

This gives us factors $(x+3+2sqrt(5))$ and $(x+3-2sqrt(5))$

So the full factorization (if we allow irrational coefficients) is:

$6x^2y+36xy-66y = 6y(x+3+2sqrt(5))(x+3-2sqrt(5))$

How do you factor 6x^2y + 36xy - 66y6x^2y + 36xy - 66y?