How do you factor completely $8 z^{3} + 27$ $8z^3 +27$ ?

Question

How do you factor completely $8 z^{3} + 27$ $8z^3 +27$ ?

1 Answer

Impact of this question

2105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-21T19:11:25

$8 x^{3} + 27 = (2 x + 3) (4 x^{2} - 6 x + 9)$ $8x^3+27=(2x+3)(4x^2-6x+9)$

Explanation:

The sum of cubes identity can be written:

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

We use this with $a = 2 z$ $a=2z$ and $b = 3$ $b=3$ to find:

$8 x^{3} + 27$ $8x^3+27$

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

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

$= (2 x + 3) (4 x^{2} - 6 x + 9)$ $=(2x+3)(4x^2-6x+9)$

Note that the remaining quadratic expression has no linear factors with Real coefficients. You can tell this from its discriminant:

$Delta = (-6)^2-4(4)(9) = 36-144 = -108 < 0$

Science

Math

Humanities

... and beyond

How do you factor completely $8 z^{3} + 27$ $8z^3 +27$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor completely 8z^3 +278z3+278z^3 +27?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor completely $8 z^{3} + 27$ $8z^3 +27$ ?