Factor completely. 40x^2 – 2xy – 24y^2?

Question

Factor completely. 40x^2 – 2xy – 24y^2

Answer 1

A square of difference is a type of quadratic equation of the form:

x^2 – 2bx + b^2

= (x – b)^2

This is gotten from the basic pattern:

x^2-2x+1
=(x-1)^2

Now, I will attempt to apply this pattern to that problem:

40x^2-2x-24
=(8x+6)(5x-4)

Then, I would assume that:

40x^2-2xy-24y^2
=(8x+6y)(5x-4y)
=2(4x+3y)(5x-4y)

I hope this works for you. I have not worked with math in over 40 years, but I was relatively the best in my class for every year when I was in school. I memorize and implement patterns very well. However, this might not work for some, I forget.

Answer 2

40x^2 – 2xy – 24y^2

2(20x^2 - xy - 12y^2)

2(5x - 4y)(4x + 3y)

For these hard trinomials, see the lesson at the link below

Answer 3

first factor out 2, this gives you 2(20x^2-xy-12y^2) then to factor the quadratic we do this:

20x^2 = 4x * 5x, and -12y^2 = -4y * 3y

2(4x + 3y)(5x - 4y)

Answer 4

2( 20 x ² - x y - 12 y ² )
2 ( 5x - 4y) (4x + 3y )

Answer 5

40x^2-2xy-24y^2
=2(20x^2-xy-12y^2)
Find two nos whose sum=-1(b) & whose product=20*(-12)=-240(ac),so we have -16 &15
=2(5x*4x-16xy+15xy-3y*4y)
=2{5x*4x-4x*4y+5x*3y-3y*4y}
=2{4x(5x-4y)+3y(5x-4y)}
=2(4x+3y)(5x-4y) ans

Answer 6

(8x + 6y)(5x - 4y)