Question

p(y) = 2y² + 5y – 3 at y=-1/2
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Answer 1

Step-by-step explanation:

2y^2+5y-3=0

2y^2+6y-y-3=0

2y(y+3)-1

(y+3)=0

(2y-1)(y+3)=0

y=1/2

y=-3 zeros of polynomial are -3,1/2.

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Answer 2

Answer:

-5

Step-by-step explanation:

p (y) = 2y^2 +5y -3

p (-1/2) = 2 (-1/2)^2 + 5 (-1/2) -3

= 2(1/4) -(5/2) -3

= 1/2 - 5/2 - 6/2

= (1 - 5 - 6)/2

= -10/2

= -5