How do you factor completely: $21x^3 + 35x^2 + 9x + 15$ ?

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Answer 1 · 2015-07-29T18:15:56

Factor by grouping to find:

$21x^3+35x^2+9x+15=(7x^2+3)(3x+5)$

$21x^3+35x^2+9x+15$

$=(21x^3+35x^2)+(9x+15)$

$=7x^2(3x+5)+3(3x+5)$

$=(7x^2+3)(3x+5)$

The term $7x^2+3$ has no linear factors with real coefficients since $7x^2+3 >= 3 > 0$ for all $x in RR$ .

How do you factor completely: 21x^3 + 35x^2 + 9x + 15 21x^3 + 35x^2 + 9x + 15 ?