Math, asked by saksham7827, 11 months ago

T-he-eht she
${2y}^{3} + {y}^{2} - 2y - 1$

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Answered by 217him217

Answer:

2y³ + y² - 2y - 1 = 0

by table method write coefficient

1 | 2. 1. -2. -1

| 0. 2. 3. 1

2. 3. 1. 0

=> (y-1)(2y²-3y+1) = 0

=> y-1 = 0 => y = 1

=> 2y²-3y+1 = 0

=> 2y² -2y - y +1=0

=>2y(y-1)-1(y-1) =0

=>(2y-1)(y-1) =0

=> y= 1/2,1

3 zeros of y = 1/2,1,1

Answered by Anonymous

Answer:

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${2y}^{3} + {y}^{2} - 2y - 1$