How do you factor $125 + 8 z^{3}$ $125+8z^3$ ?

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Answer 1 · 2015-12-24T05:59:13

$(5 + 2 z) (25 - 10 z + 4 z^{2})$ $(5+2z)(25-10z+4z^2)$ or
$(2 z + 5) (4 z^{2} - 10 z + 25)$ $(2z+ 5)(4z^2 -10z+25)$

This is a sum of cube because

$125 = 5^{3}$ $125 = 5^3$

$8 z^{3} = {(2 z)}^{3}$ $8z^3 = (2z)^3$

Remember the formula for sum of cube is

$x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ $x^3 + y^3 = (x+y)(x^2 -xy+y^2)$

We can rewrite $125 + 8 z^{3} = 5^{3} + {(2 z)}^{3}$ $125 + 8z^3 = 5^3 + (2z)^3$

This factor to

$(5 + 2 z) (25 - 10 z + 4 z^{2})$ $(5 + 2z)(25-10z + 4z^2)$

or we can rewrite the factor as

$(2 z + 5) (4 z^{2} - 10 z + 25)$ $(2z + 5) (4z^2 -10z+25)$ so it can be in alphabetical order (standard form for polynomial)

How do you factor 125+8z^3125+8z3125+8z^3?