Question

solve it...........​

Answer 1

ANSWER:

• Value of P(y) at y = 0 is 3.

GIVEN:

• P(y) = 5y-4y²+3

TO FIND:

• Value of above expression at y = 0.

SOLUTION:

=> P(y) = 5y-4y²+3

Finding P(y) at y = 0.

=> P(0) = 5(0)-4(0)²+3

=> P(0) = 0-0+3

=> P(0) = 3

Value of P(y) at y = 0 is 3.

NOTE:

Some important formulas:

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab

Answer 2

Answer:

-4ysq + 5y +3

putting value of y

-4(0) sq + 5 (0) +3

0+0 +3

3

value of the polynomial is 3

Hope the answer will help

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