factor comun por agrupamiento 2m4+m3-16m-8 tambien este otro 4p3+12p2a-pa2-3a3 otro (b2-3c)2-(b2-3c) (16-3c)?

Question

ayudenme para ir entendiendo gracias

Answer 1

1.- 2m4+ m3-16m-8= m³ (2m+1) - 8 (2m +1)= (2m+1)(m³-8)

2.-
4p3+ 12p2a-pa2-3a3=
4p² ( p+3a) - a ( p+3a) = (p+3a) (4p²-a²)

3.-
estas segura que es así?
si es asi, resuelves:
(b²-3c) - (b²-3c) ( 16-3c)=
(b²-3c) [ (b²-3c) - (16-3c)] =
(b²-3c) [b²-3c-16+3c] =(b²-3c) ( b² -16)

Suerte!!!

Answer 2

2m^4+m³-16m-8
m³(2m + 1) - 8(2m + 1) = 0
(m³ - 8)(2m + 1) = 0
m³ = 8
m = ³\/8
m = 2
2m + 1
2m = -1
m = -1/2
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4p³+12p²a-pa²-3a³ = 0
4p²(p + 3a) - a²(p + 3a) = 0
(4p² - a²) (p + 3a) = 0
(2p - a)(2p + a)(p + 3a) = 0
2p = a
p = a/2
2p = -a
p = -a/2
p = -3a
><
(b²-3c)²-(b²-3c) (16-3c) = 0
-3c = -b
c = b²/3

-3c = -16
c = 16/3
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Answer 3

2m4+m3-16m-8 = m3(1+2m)-8(1+2m)= (m3-8)(1+2m)

(b2-3c)2-(b2-3c)(16-3c)= (b2-3c)(-16+3c+2=(b2-3c)(-14+3c)