SOLUTION: factor: 2x^5 + x^4 - 8x^3 - 4x^2

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Question 209557: factor: 2x^5 + x^4 - 8x^3 - 4x^2
Found 3 solutions by stanbon, nerdybill, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
factor: 2x^5 + x^4 - 8x^3 - 4x^2
----
x^2(2x^3 + x^2 - 8x - 4)
---
I graphed the cubic and found a zero at x = 2
Then using synthetic division I got:
2)....2.....1.....-8.....-4
.......2.....5......2....|..0
Quotient: 2x^2+5x+2 = (2x+1)(x+2)

= x^2(2x+1)(x-2)(x+2)
==============================
Cheers,
Stan H.
Reply to stanbon@comcast.net

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
factor:
2x^5 + x^4 - 8x^3 - 4x^2
.
Begin by factoring out what is common among all terms:
x[2x^4 + x^3 - 8x^2 - 4x]
.
Group terms:
x[(2x^4 + x^3) - (8x^2 + 4x)]
.
Factor expressions inside parentheses:
x[x^3(2x + 1) - 4x(2x + 1)]
x[(2x + 1)(x^3- 4x)]
x(2x + 1)(x^3- 4x)

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E5+%2B+x%5E4+-+8x%5E3+-+4x%5E2
x%5E2%282x%5E3+%2B+x%5E2+-+8x+-+4%29 ------ Factor out GCF, x%5E2
x%5E2%28x%5E2%282x+%2B+1%29+-+4%282x+%2B+1%29%29 ------ Factor the inner polynomial

Therefore, we have: x%5E2%28x%5E2+-+4%29%282x+%2B+1%29

Factor x%5E2++-+4 to get: highlight_green%28x%5E2%28x+-+2%29%28x+%2B+2%29%282x+%2B+1%29%29