How do you factor $x^{5} + x^{3} + 8 x^{2} + 8$ $x^5+x^3+8x^2+8$ ?

Question

How do you factor $x^{5} + x^{3} + 8 x^{2} + 8$ $x^5+x^3+8x^2+8$ ?

1 Answer

Impact of this question

4962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-16T21:53:55

Factor by grouping and the sum of cubes identity to find:

$x^{5} + x^{3} + 8 x^{2} + 8 = (x^{2} + 1) (x + 2) (x^{2} - 2 x + 4)$ $x^5+x^3+8x^2+8 = (x^2+1)(x+2)(x^2-2x+4)$

Explanation:

The sum of cubes identity may 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 $a = x$ $a=x$ and $b = 2$ $b=2$ to find:

$x^{5} + x^{3} + 8 x^{2} + 8$ $x^5+x^3+8x^2+8$

$= (x^{5} + x^{3}) + (8 x^{2} + 8)$ $=(x^5+x^3)+(8x^2+8)$

$= (x^{2} + 1) x^{3} + (x^{2} + 1) 8$ $=(x^2+1)x^3 + (x^2+1)8$

$= (x^{2} + 1) (x^{3} + 8)$ $=(x^2+1)(x^3+8)$

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

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

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

Science

Math

Humanities

... and beyond

How do you factor $x^{5} + x^{3} + 8 x^{2} + 8$ $x^5+x^3+8x^2+8$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor x^5+x^3+8x^2+8x5+x3+8x2+8x^5+x^3+8x^2+8?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $x^{5} + x^{3} + 8 x^{2} + 8$ $x^5+x^3+8x^2+8$ ?