#1

yea i know math thread, can never get replies in there. i'll delete this one as soon as i get a good response.

anyways

its a question about factoring which i seem to be completely blanking on.

how would i factor an equation such as

28x^8y^3+10x^5y^9

also if you factored

x^3

would it just be (x)(x)(x)

anyways

its a question about factoring which i seem to be completely blanking on.

how would i factor an equation such as

28x^8y^3+10x^5y^9

also if you factored

x^3

would it just be (x)(x)(x)

#2

28*x^8*y^3+10*x^5*y^9 =

= 2 * x^5 * y^3 * (14*x^3 + 5*y^6 )

and its too late for me to bother to figure out if you can factor whats left in the brackets

iirc you can factor the sum of cubes

edit: a^3 + b^3 = (a + b)(a^2 - a*b + b^2).

apply that

y^6 = (y^2)^3

(14*x^3 + 5*y^6 ) =

= (14^(1/3) * x)^3 + (5^(1/3) * y^2)^3 =

= (14^(1/3) * x + 5^(1/3) * y^2) * (14^(2/3) * x^2 - 70^(1/3)*x*y^2 + 5^(2/3)*y^4)

so altogether

28*x^8*y^3+10*x^5*y^9 =

= 2 * x^5 * y^3 * (14*x^3 + 5*y^6 )

and its too late for me to bother to figure out if you can factor whats left in the brackets

iirc you can factor the sum of cubes

edit: a^3 + b^3 = (a + b)(a^2 - a*b + b^2).

apply that

y^6 = (y^2)^3

(14*x^3 + 5*y^6 ) =

= (14^(1/3) * x)^3 + (5^(1/3) * y^2)^3 =

= (14^(1/3) * x + 5^(1/3) * y^2) * (14^(2/3) * x^2 - 70^(1/3)*x*y^2 + 5^(2/3)*y^4)

so altogether

28*x^8*y^3+10*x^5*y^9 =

**= 2 * x^5 * y^3 * (14^(1/3) * x + 5^(1/3) * y^2) * (14^(2/3) * x^2 - 70^(1/3)*x*y^2 + 5^(2/3)*y^4)***Last edited by seljer at Sep 29, 2008,*

#3

You can pull out an x^5 and y^3 out of both terms and you get:

x^5 * y^3 * (28x^3 + 10y^6)

Now pull out the largest common factor between 28 and 10 and get:

2 * x^5 * y^3 * (14x^3 + 5y^6)

x^5 * y^3 * (28x^3 + 10y^6)

Now pull out the largest common factor between 28 and 10 and get:

2 * x^5 * y^3 * (14x^3 + 5y^6)