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B
False
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Solution
The correct option is A True (2y−4y3+6y5)×(y2+y−3) =(2y−4y3+6y5)y2+(2y−4y3+6y5)y−3(2y−4y3+6y5) =2y3−4y5+6y7+2y2−4y4+6y6−6y+12y3−18y5 =6y7+6y6−22y5−4y4+14y3+2y2−6y