How do you factor $18 x^{3} + 9 x^{5} - 27 x^{2}$ $18x^3+9x^5-27x^2$ ?

Question

How do you factor $18 x^{3} + 9 x^{5} - 27 x^{2}$ $18x^3+9x^5-27x^2$ ?

1 Answer

Impact of this question

2087 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-22T20:21:42

Find: $18 x^{3} + 9 x^{5} - 27 x^{2} = 9 x^{2} (x - 1) (x^{2} + x + 3)$ $18x^3+9x^5-27x^2=9x^2(x-1)(x^2+x+3)$

as shown below...

Explanation:

Rearrange in standard order (descending powers of $x$ $x$ ) and separate out the common factor $9 x^{2}$ $9x^2$ which all the terms are divisible by:

$18 x^{3} + 9 x^{5} - 27 x^{2}$ $18x^3+9x^5-27x^2$

$= 9 x^{5} + 18 x^{3} - 27 x^{2}$ $=9x^5+18x^3-27x^2$

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

Next note that the sum of the coefficients of $x^{3} + 2 x - 3$ $x^3+2x-3$ is $0$ $0$ , so $x = 1$ $x=1$ is a zero of this cubic and $(x - 1)$ $(x-1)$ is a factor:

$= 9 x^{2} (x - 1) (x^{2} + x + 3)$ $=9x^2(x-1)(x^2+x+3)$

The discriminant of the remaining quadratic factor is $1^{2} - (4 \times 1 \times 3) = - 11$ $1^2-(4xx1xx3) = -11$ which is negative, so there are no simpler factors with Real coefficients.

If you still want to factor it further you can use Complex coefficients:

$(x^{2} + x + 3) = {(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}$ $(x^2+x+3) = (x+1/2)^2+(sqrt(11)/2)^2$

$= (x + \frac{1}{2} - \frac{\sqrt{11}}{2} i) (x + \frac{1}{2} + \frac{\sqrt{11}}{2} i)$ $= (x+1/2-sqrt(11)/2 i)(x+1/2+sqrt(11)/2 i)$

Science

Math

Humanities

... and beyond

How do you factor $18 x^{3} + 9 x^{5} - 27 x^{2}$ $18x^3+9x^5-27x^2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor 18x3+9x5−27x218x^3+9x^5-27x^2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $18 x^{3} + 9 x^{5} - 27 x^{2}$ $18x^3+9x^5-27x^2$ ?