Algebra 2 factoring?

a) 3x^3 + 192
Sum of cubes, with a factor of 3.
(a³ + b³) = (a + b)(a² - ab + b²)

3x^3 + 192 =
3(x³ + 64) =
3( (x)³ + (4)³ ) =
3( x + 4 )( x² - 4x + 16)

b) 8x^3 - 64
Difference of cubes
(a³ - b³) = (a - b)(a² + ab + b²)

8x³ - 64 =
8(x³ - 8) =
8( (x)³ - (2)³ ) =
8( x - 2 )( x² + 2x + 4)

c) 4x^3 - 6x^2 + 10x - 15
Note the first two terms are divisible by 2x², and the last 2 terms are divisible by 5. This usually leads to being able to factor by grouping:
2x² (2x - 3) + 5(2x - 3) =
(2x² + 5)(2x - 3)

d) 9x^3 + 18x^2 + 7x + 14
Note the first two terms are divisible by 9x², and the last 2 terms are divisible by 7. Again, factor by grouping:
9x²(x + 2) + 7(x + 2) =
(9x² + 7)(x + 2)

a) 3(x^3 + 64)

b) 8(x^3 - 8)

c) 2x(2x^2 - 3x + 5) - 15

d) 9x^2(x + 2) + 7(x + 2)
(x + 2)( 9x^2 +7)

a)
3x^2 + 192
= 3(x^2 + 64)

b)
8x^3 - 64
= (2x - 4)(4x^2 + 8x + 16)

c)
4x^3 - 6x^2 + 10x - 15
= (2x^2 + 5)(2x - 3)

d)
9x^3 + 18x^2 + 7x + 14
= (9x^2 + 7)(x + 2)

a) 3x^3 + 192=3(x^3+64)=3(x+4)(x^2-4x+16)

b) 8x^3 - 64=8(x^3-8)=8(x-2)(x^2+2x+4)

c) 4x^3 - 6x^2 + 10x - 15=2x^2(2x-3)+5(2x-3)=(2x-3)(2x^2+5)

d) 9x^3 + 18x^2 + 7x + 14=9x^2(x+2)+7(x+2)=(x+2)(9x^2+7)

b)8x^3=64
x^3=8
x=2 since 2X2X2=8