You can put this solution on YOUR website! factor: 2x^5 + x^4 - 8x^3 - 4x^2
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x^2(2x^3 + x^2 - 8x - 4)
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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)
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Cheers,
Stan H.
Reply to stanbon@comcast.net
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)