factor the trinomial completely: 14y^4 + 94y^3 - 28y^2?
回答 (4)
2015-09-03 7:16 am
2y^2(y+7)(7y-2)
2015-09-01 5:01 pm
y^2(14y^2+94y-28)
y^2(14y^2+98y-4y-28)
y^2(14y(y+7)-4(y+7)
y^2(y+7)(14y-4) or if you prefer...
2y^2(y+7)(7y-2)
y^2(14y^2+98y-4y-28)
y^2(14y(y+7)-4(y+7)
y^2(y+7)(14y-4) or if you prefer...
2y^2(y+7)(7y-2)
2015-09-01 4:35 pm
14y⁴ + 94y³ - 28y²
= 2y²(7y² + 47y - 14)
= 2y²(7y² - 2)(y + 7)
= 2y²(7y² + 47y - 14)
= 2y²(7y² - 2)(y + 7)
2015-09-01 4:31 pm
2*y^2*(7*y^2 + 47*y - 14)
2*y^2*(7*y - 2)*(y + 7) <<<
2*y^2*(7*y - 2)*(y + 7) <<<
收錄日期: 2021-05-01 13:04:55
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