Factor completely as a product of prime polynomials.
255(x+3y)^8+49 a^4b^12
^ this means to # power .. 8th power, 4th power,12th power
thank you
2007-02-17 07:29:28 · 5 answers · asked by abc 1 in Science & Mathematics ➔ Mathematics
255(x+3y)^8+49 a^4b^12
That seems perfectly satisfactory to me. Any other rearrangement is likely to be reduced to this one.
HTH
Charles
2007-02-17 07:40:33 · answer #1 · answered by Charles 6 · 0⤊ 0⤋
I'm thinking that this is factored completely. We can express it as a sum of squares: 255((x+3y)^4)^2+(7a^2b^6)^2 ... and the sum of squares cannot be factored.
2007-02-17 15:46:43 · answer #2 · answered by koolkat 3 · 0⤊ 0⤋
this is the sum of two squares... I don't think that factors.
c^2 + d^2 doesn't factor any further, right?
c= 255^.5 (x+3y)^4
d= 7 (a)^2 (b)^6
2007-02-17 15:46:32 · answer #3 · answered by David T 4 · 0⤊ 0⤋
255(x+3y)^8+49 a^4 b^12
{15[(x+3y)^2]^2+7a^2 [(b)^3]^2}^2
2007-02-17 15:52:25 · answer #4 · answered by Muhammad Faraz Quadri 2 · 0⤊ 0⤋
Are you sure this isn't 256 instead of 255 and - instead of + ?
If it is, then the answer is
[16(x+3y)^4+7a^2b^6] * [16(x+3y)^4-7a^2b^6]
see your other question for the steps
2007-02-17 15:41:05 · answer #5 · answered by andthendougsaid 2 · 0⤊ 0⤋