How do you factor #6x^{2}-13xy+5y^{2}#?

Question

2551 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-14T06:43:27

#6x^2-13xy+5y^2 = (2x-y)(3x-5y)#

Given:

#6x^2-13xy+5y^2#

Note that this is a homogeneous polynomial in that all of the terms are of the same degree (#2#).

As a result, we can use methods similar to those we would use with a quadratic in one variable.

Putting #a=6#, #b=-13# and #c=5# we find that the discriminant is:

#Delta = b^2-4ac = (color(blue)(-13))^2-4(color(blue)(6))(color(blue)(5)) = 169-120 = 49 = 7^2#

Since this is positive and a perfect square, we can tell that this quadratic will factor with integer coefficients.

Let's use an AC method:

Find a pair of factors of #AC = 6*5=30# with sum #B=13#.

The pair #10#, #3# works in that #10 * 3 = 30# and #10+3=13#.

Use this pair to split the middle term and factor by grouping:

#6x^2-13xy+5y^2 = (6x^2-10xy)-(3xy-5y^2)#

#color(white)(6x^2-13xy+5y^2) = 2x(3x-5y)-y(3x-5y)#

#color(white)(6x^2-13xy+5y^2) = (2x-y)(3x-5y)#