pls.... help me about factoring trinomials...

Question

anonymous · Answer

$$4a ^{2}b ^{2}+12abc+9c ^{2}$$

anonymous · Answer

Notice 2ab times 2ab = 4a^2b^2
Notice 3c times 3c = 9c^2

Therefore we have

(2ab + 3c)(2ab + 3c)

anonymous · Answer

how about this...$$16a ^{6}m ^{2}-8a ^{4}mn ^{2}+a ^{2}n ^{4}$$

anonymous · Answer

Notice that we have a GCF of a^2, take this out

a^2(16a^4m^2 - 8a^2mn^2 + n^4)

Then notice
4a^2m times 4a^2m = 16a^4m^2
and
n^2 times n^2 = n^4

Therefore

a^2(4a^2m + n^2)(4a^2m + n^2)

anonymous · Answer

$$\left( x-y \right)^{5}-243z ^{^{5}}$$

anonymous · Answer

help me pls..

anonymous · Answer

Using Technology:

$$ (x-y-3 z) \left(x^4-4 x^3 y+6 x^2 y^2-4 x y^3+y^4+3 x^3 z-9 x^2 y z+9 x y^2 z-3 y^3 z+9 x^2 z^2-18 x y z^2+9 y^2 z^2+27 x z^3-27 y z^3+81 z^4\right) $$

anonymous · Answer

$$32a ^{10}b ^{15}-243c ^{10}$$

anonymous · Answer

pls....

anonymous · Answer

$$32a ^{10}b ^{15}-243c ^{10}$$

anonymous · Answer

$$ \left(2 a^2 b^3-3 c^2\right) \left(16 a^8 b^{12}+24 a^6 b^9 c^2+36 a^4 b^6 c^4+54 a^2 b^3 c^6+81 c^8\right) $$

anonymous · Answer

thank you..