Factor 18x^2 +24xy-10y^2 completely.?

Answer 1

Factor out what is common in this equation first.
Then do the backwards FOIL method.

= 2 (9x^2 + 12 xy - 5y^2)
= 2 (3x - y)(3x + 5y)

Answer 2

Factor out 2 to start --> 18x^2 + 24 xy -10y^2 = 2*[9x^2 +12 xy - 5 y^2] = 2*((3x+5)(3x-1))

Answer 3

First step is ALWAYS to factor out a common term.
18, 24, & 10 are all divisible by 2


2[9x^2+12xy-5y^2]

use the ac method to find two factors that multiply to -45 and add up to +12
15*(-3) =-45
15+(-3) =+12

2[9x^2+15xy-3xy-5y^2] replace +12xy with +15xy-3xy
2[(9x^2+15xy)-(3xy+5y^2)]factor by grouping
2[3x(3x+5y)-y(3x+5y)]factor
2(3x+5y)[3x-y]factor out a common term
2(3x+5)(3x-y) solution

Answer 4

ok: simplify to

2(9x^2 +12xy - 5y^2)

factor what's in the brackets, lets start with 9x^2, and guess even factors of 3x

so: (3x )(3x ), now we know the last terms must multiply to give 5y^2, and as 5 is prime we know the last terms are y and 5y so:

(3x 1y)(3x 5y) now we just need the signs, knowing that one is negative and one positive (as the term is -5y^2)...so using FOIL

if we expand with -y and +5y we get

(3x-y)(3x+5y)= 9x^2 +15xy -3xy - 5y^2)=(9x^2 +12xy -5y^2)

Which is what we are after....so fully factored it is:

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

hope that the explaination makes sense to you!

Answer 5

18x^2 + 24xy - 10y^2
= 2 (9x^2 + 12xy - 5y^2)
= 2 (9x^2 - 3xy + 15xy - 5y^2)
= 2 [3x (3x - y) + 5y (3x - y) ]
= 2 (3x-y) (3x + 5y)

Answer 6

2(9x^2+12xy-5y^2)
2(3x+5y)(3x-1y)