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

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Answer 1 · 2015-07-29T18:15:56

Factor by grouping to find:

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

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

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

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

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

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

How do you factor completely: 21x^3 + 35x^2 + 9x + 15 21x3+35x2+9x+1521x^3 + 35x^2 + 9x + 15 ?