What is factor 3x^3+23x^2+40x-16 ?

The solution to factor 3x^3+23x^2+40x-16 is (x+4)^2(3x-1)

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factor 3x^3+23x^2+40x-16

factor

3 x^{3} + 23 x^{2} + 40 x - 16

(x + 4)^{2} (3 x - 1)

Solution steps

3 x^{3} + 23 x^{2} + 40 x - 16

Use the rational root theorem

= (x + 4) \frac{3 x ^{3} + 23 x ^{2} + 40 x - 16}{x + 4}

\frac{3 x ^{3} + 23 x ^{2} + 40 x - 16}{x + 4} = 3 x^{2} + 11 x - 4

= (x + 4) (3 x^{2} + 11 x - 4)

Factor

3 x^{2} + 11 x - 4 : (3 x - 1) (x + 4)

= (x + 4) (3 x - 1) (x + 4)

Apply exponent rule:

aa = a^{2}

(x + 4) (x + 4) = (x + 4)^{2}

= (x + 4)^{2} (3 x - 1)

- 5.3 \geq 6.7 + 4.3 + q

2 x^{3} + 7 x^{2} - 10 x - 24

(- 4 x^{4} y^{4})^{3}

∣ x + 4 ∣ > - 1

4 x^{2} + 19 x + 21

What is factor 3x^3+23x^2+40x-16 ?
The solution to factor 3x^3+23x^2+40x-16 is (x+4)^2(3x-1)