divide 2y^3+y^2-2y-1 by y+1
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Answered by
7
2y³ + y² - 2y -1 = (2y³ + y²) - (2y +1)
= y²(2y + 1) - 1x(2y+1)
=(2y+1)(y² - 1)
= (2y+1)(y+1)(y-1)
Henve dividing (2y³ + y² - 2y -1) / (y+1)
= (2y+1)(y+1)(y-1) / (y+1)
= (2y+1)(y-1)
= y²(2y + 1) - 1x(2y+1)
=(2y+1)(y² - 1)
= (2y+1)(y+1)(y-1)
Henve dividing (2y³ + y² - 2y -1) / (y+1)
= (2y+1)(y+1)(y-1) / (y+1)
= (2y+1)(y-1)
Answered by
3
2y³ + y² - 2y -1
= (2y³ + y²) - (2y +1)
= y²(2y + 1) - 1x(2y+1)
=(2y+1)(y² - 1)
= (2y+1)(y+1)(y-1)
Hence dividing, (2y³ + y² - 2y -1) / (y+1)
= (2y+1)(y+1)(y-1) / (y+1)
= (2y+1)(y-1)
= (2y³ + y²) - (2y +1)
= y²(2y + 1) - 1x(2y+1)
=(2y+1)(y² - 1)
= (2y+1)(y+1)(y-1)
Hence dividing, (2y³ + y² - 2y -1) / (y+1)
= (2y+1)(y+1)(y-1) / (y+1)
= (2y+1)(y-1)
Anonymous:
I hope it will help you.
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