How do you factor $y= 8t^4+ 32t ^3 + t + 4$ ?

Question

1678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-24T05:42:36

$y=(4t^2-2t+1)(2t+1)(t+4)$

Factor by grouping. (Split into two groups of two.)

$y=(8t^4+32t^3)+(t+4)$

$y=8t^3(t+4)+1(t+4)$

Factor out a common $(t+4)$ .

$y=(8t^3+1)(t+4)$

Recognize that $(8t^3+1)$ is a sum of cubes.

This is the following identity:

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

Thus, $a=2t,b=1$ , so

$8t^3+1=(2t+1)(4t^2-2t+1)$

Replace this in the factored equation:

$y=(4t^2-2t+1)(2t+1)(t+4)$

How do you factor y= 8t^4+ 32t ^3 + t + 4 y= 8t^4+ 32t ^3 + t + 4 ?