factoring

[Guest]

i was wondering if someone could help me factor this:

a^2+4b^2-25c^2+4ab

since there are 4 or more terms i know you have to group them but i dont see anything that the -25c^2 could be grouped with?

and also i have another one..

1000f^3 +27^3

i dont see a cube patten since 1000 is not a cube and there is no GCF so i dont know what to do?

 

[rahidz2003]

For the second one, 1000f^3 = (10f)^3, and 19683 = 27^3, so if we let a=10f and b=27, it factors as

(a^3+b^3)=

(a+b)(a^2+b^2-ab)=

(10f + 27) * (100f^2 + 729 - 270f)

 

[stapel]

When they give you four terms, and nothing comes out of everything, try to factor "in pairs". For instance:

. . . . .x³ + 3x² - 6x - 18

. . . . .(x³ + 3x²) + (-6x - 18)

. . . . .x²(x + 3) - 6(x + 3)

. . . . .(x + 3)(x² - 6)

However, that won't work in this case, because there's only one term with the variable "c" in it (so there's nothing to pair it with). So see if you can find a squared binomial:

. . . . .a² + 4b² - 25c² + 4ab

. . . . .a² + 4ab + 4b² - 25c²

. . . . .(a² + 4ab + 4b²) - (5c)²

You should be able now to see a way to convert this to a difference of squares, and then factor using that formula.

Your second expression is just a sum of cubes, so apply the appropriate formula. (I don't know who told you that one thousand is not a cube, but s/he was wrong.)

Eliz.

 

[Guest]

For the second one, 1000f^3 = (10f)^3, and 19683 = 27^3, so if we let a=10f and b=27, it factors as

(a^3+b^3)=

(a+b)(a^2+b^2-ab)=

(10f + 27) * (100f^2 + 729 - 270f)

on the second one, i'm confused because don't you let
a=10f
b=3 because 3^3 is 27 making the answer

(10f+3)(100f2-30f+9 ? from the formula (a+b)(a^2-ab+b^2)

 

[Gene]

The original problem had 27^3 so the cube root is (surprise) 27. It is just a coincidence that 27=3^3.
b=27