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4x^3-32 need to factor
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take 4 common and solve
so it becomes 4 (x^3-8)
4(x^3 -2^3)
now 8 can be written as 2^3 ....just simply and use a^3-b^3
\( \color{Black}{\Rightarrow 4(x^2 - 8) }\) can be factored(difference of cubes)
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4(x^3 - 8)*
\( \color{Black}{\Rightarrow a^3 - b^ 3 = (a - b)(a^2 + ab + b^2) }\)
the both of you guys really confused me here
1. Take the gcf of 4 out. \[\rightarrow 4(x^3 - 8)\] 2. Since 8 = 2^2, use the difference of squares (@ParthKohli gives the form correctly) to factor it.\[\rightarrow 4(x-2)(x^2 +2x+4)\]That is as factored as you can make it.
thanks
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