find the zeroes of each polynomial 21y2-11y- 2
Answers
Answered by
23
Answer:
2/3 and -1/7
Step-by-step explanation:
given :- 21y² - 11y - 2
= 21y² - (14y - 3y) - 2
= 21y² - 14y + 3y - 2
= 7y(3y - 2) + 1(3y - 2)
= (3y - 2) (7y + 1)
equating with zero
» 3y - 2 = 0 » 3y = 2 » y = 2/3
» 7y + 1 = 0 » 7y = -1 » y = -1/7
hence, the zeroes of the polynomial are 2/3 and -1/7
Answered by
9
Answer:
2/3 and -1/7
Step-by-step explanation:
given :- 21y² - 11y - 2
= 21y² - (14y - 3y) - 2
= 21y² - 14y + 3y - 2
= 7y(3y - 2) + 1(3y - 2)
= (3y - 2) (7y + 1)
equating with zero
» 3y - 2 = 0 » 3y = 2 » y = 2/3
» 7y + 1 = 0 » 7y = -1 » y = -1/7
hence, the zeroes of the polynomial are 2/3 and -1/7
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