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jmc

algebra intermediate

Problem

Factor

58 x^{5} - 203 x^{11}

.

Solution

Since

58 = 2 \cdot 29

and

203 = 7 \cdot 29

, we can factor a

29 x^{5}

from the expression, to get

58 x^{5} - 203 x^{11} = - 29 x^{5} (7 x^{6} - 2) .

Final answer

-29x^5(7x^6-2)