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

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Answer 1 · 2015-12-24T05:42:36

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

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

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

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

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

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

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

This is the following identity:

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

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

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

Replace this in the factored equation:

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

How do you factor y=8t4+32t3+t+4y= 8t^4+ 32t ^3 + t + 4 ?