How do you factor $6 x^{3} + 48$ $6x^3+48$ ?

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Answer 1 · 2015-12-18T19:45:06

Separate out the scalar factor $6$ $6$ then use the sum of cubes identity to find:

$6 x^{3} + 48 = 6 (x + 2) (x^{2} - 2 x + 4)$ $6x^3+48 = 6(x+2)(x^2-2x+4)$

First separate out the common scalar factor $6$ $6$ to find:

$6 x^{3} + 48 = 6 (x^{3} + 8)$ $6x^3+48 = 6(x^3+8)$

Then notice that both $x^{3}$ $x^3$ and $8 = 2^{3}$ $8 = 2^3$ are perfect cubes, so work well with the sum of cubes identity:

$A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})$ $A^3+B^3 = (A+B)(A^2-AB+B^2)$

With $A = x$ $A=x$ and $B = 2$ $B=2$ we find:

$x^{3} + 8 = (x^{2} + 2^{3}) = (x + 2) (x^{2} - 2 x + 4)$ $x^3+8 = (x^2+2^3) = (x+2)(x^2-2x+4)$

Putting it together we get:

$6 x^{3} + 48 = 6 (x + 2) (x^{2} - 2 x + 4)$ $6x^3+48 = 6(x+2)(x^2-2x+4)$

This has no simpler factors with Real coefficients, as you can check by looking at the discriminant $Delta$ of $(x^2-2x+4)$

$Delta = b^2-4ac = (-2)^2-(4xx1xx4) = 4-16 = -12$

Since $Delta < 0$ this quadratic has no Real zeros and no linear factors with Real coefficients.

Answer 2 · 2015-12-18T19:47:17

First factor out the 6 ...

$6(x^3+8)$

Now use the identity for the sum of cubes ...

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$6(x+2)(x^2-2x+4)$

hope that helped

How do you factor 6x^3+486x3+48 6x^3+48?