Question #fc138

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Answer 1 · 2016-11-30T18:33:49

$k^{3} + 3 k^{2} - 25 k - 75 = (k + 3) (k - 5) (k + 5)$ $k^3+3k^2-25k-75=(k+3)(k-5)(k+5)$

Factor by grouping first

$k^{3} + 3 k^{2} - 25 k - 75 = k^{2} (k + 3) - 25 (k + 3)$ $k^3+3k^2-25k-75=k^2(k+3)-25(k+3)$

factor out the common binomial factor $(k + 3)$ $(k+3)$

$k^{2} (k + 3) - 25 (k + 3) = (k + 3) (k^{2} - 25)$ $k^2(k+3)-25(k+3)=(k+3)(k^2-25)$

then factor the
difference of two squares $k^{2} - 25 = (k - 5) (k + 5)$ $k^2-25=(k-5)(k+5)$

then continue

$(k + 3) (k^{2} - 25) = (k + 3) (k - 5) (k + 5)$ $(k+3)(k^2-25)=(k+3)(k-5)(k+5)$

the complete factored form is

$k^{3} + 3 k^{2} - 25 k - 75 = (k + 3) (k - 5) (k + 5)$ $k^3+3k^2-25k-75=(k+3)(k-5)(k+5)$

God bless....I hope the explanation is useful.