how can i factor
1.x^3+1/8
2.) 5x^3+40y^6
my answer is
(4x^4-2xy^2+x)2x
but this is part right.now correct
2007-07-25 15:30:55 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1) x^3+1/8
= (x+1/2)(x^2 - 1/2x +1/4)
2) 5x^3+40y^6
= 5(x^3 + 8y^6)
= 5(x+2y^2)(x^2 - 2xy^2 + 4y^4)
In general,
a^3 + b^3
= (a+b)(a^2 - ab + b^2)
I hope this helps!
2007-07-25 15:36:45 · answer #1 · answered by math guy 6 · 0⤊ 0⤋
1) x^3+1/8
= (x+1/2)(x^2 - 1/2x +1/4)
2) 5x^3+40y^6
= 5(x^3 + 8y^6)
= 5(x+2y^2)(x^2 - 2xy^2 + 4y^4)
we used the formula,
a^3 + b^3
= (a+b)(a^2 - ab + b^2)
2007-07-25 22:52:03 · answer #2 · answered by Sumita T 3 · 0⤊ 0⤋
1.) x^3+1/8
= (x+1/2)(x^2 - x/2 +1/4)
2.) 5x^3+40y^6
= 5(x^3 +8y^6)
= 5(x+2y^2)(x^2 -2xy^2 +4y^4)
2007-07-25 15:55:29 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
1) (x+1/2)^3
2) 5[(x+2y^2)^3]
2007-07-25 15:36:49 · answer #4 · answered by i am this 2 · 0⤊ 1⤋
Do you have a teacher? You should ask him/her for help.
2007-07-25 15:33:59 · answer #5 · answered by M.G.G. 2 · 0⤊ 1⤋