Factor 12x^3+31x^2+7x

$12 x^{3} + 31 x^{2} + 7 x$

Step 1

Factor $x$ out of $12 x^{3} + 31 x^{2} + 7 x$ .

Factor $x$ out of $12 x^{3}$ .

$x (12 x^{2}) + 31 x^{2} + 7 x$

Factor $x$ out of $31 x^{2}$ .

$x (12 x^{2}) + x (31 x) + 7 x$

Factor $x$ out of $7 x$ .

$x (12 x^{2}) + x (31 x) + x \cdot 7$

Factor $x$ out of $x (12 x^{2}) + x (31 x)$ .

$x (12 x^{2} + 31 x) + x \cdot 7$

Factor $x$ out of $x (12 x^{2} + 31 x) + x \cdot 7$ .

$x (12 x^{2} + 31 x + 7)$

Step 2

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 12 \cdot 7 = 84$ and whose sum is $b = 31$ .

Factor $31$ out of $31 x$ .

$x (12 x^{2} + 31 (x) + 7)$

Rewrite $31$ as $3$ plus $28$

$x (12 x^{2} + (3 + 28) x + 7)$

Apply the distributive property.

$x (12 x^{2} + 3 x + 28 x + 7)$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x ((12 x^{2} + 3 x) + 28 x + 7)$

Factor out the greatest common factor (GCF) from each group.

$x (3 x (4 x + 1) + 7 (4 x + 1))$

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

$x ((4 x + 1) (3 x + 7))$

Remove unnecessary parentheses.

$x (4 x + 1) (3 x + 7)$

Factor 12x^3+31x^2+7x

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