My question is to please help me with factoring completly. here some that i need help on just three? 1. #x^3-8# 2. #27^6+125y^3# 3. #x^3-2x^2-4x+8#

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I don't really understand much especially gcf and stuff like factoring completely please help,

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Answer 1 · 2016-07-11T20:31:03

1) #x^3-8 = (x-2)(x^2+2x+4)#

2) #27^6+125y^3 = (729+5y)(531441-3645y+25y^2)#

#27x^6+125y^3 = (3x^2+5y)(9x^4-15x^2y+25y^2)#

3) #x^3-2x^2-4x+8 = (x-2)(x+2)(x-2)#

The difference of cubes identity can be written:

#a^3-b^3=(a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

The difference of squares identity can be written:

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

#color(white)()#
Question 1

Use the difference of cubes identity with #a=x# and #b=2#:

#x^3-8#

#=x^3-2^3#

#=(x-2)(x^2+2x+2^2)#

#=(x-2)(x^2+2x+4)#

#color(white)()#
Question 2

Use the sum of cubes identity with #a=27^2 = 729# and #b=5y#:

#27^6+125y^3#

#=(27^2)^3+(5y)^3#

#=(729+5y)(729^2-729*5y+(5y)^2)#

#=(729+5y)(531441-3645y+25y^2)#

Actually I suspect a typo in the question. If #27^6# was supposed to be #27x^6#, then use the sum of cubes identity with #a=3x^2# and #b=5y# to find:

#27x^6+125y^3#

#=(3x^2)^3+(5y)^3#

#=(3x^2+5y)((3x^2)^2-(3x^2)(5y)+(5y)^2)#

#=(3x^2+5y)(9x^4-15x^2y+25y^2)#

#color(white)()#
Question 3

Factor by grouping:

#x^3-2x^2-4x+8#

#=(x^3-2x^2)-(4x-8)#

#=x^2(x-2)-4(x-2)#

#=(x^2-4)(x-2)#

#=(x^2-2^2)(x-2)#

#=(x-2)(x+2)(x-2)#