Question

find the zeroes 3y^2-8+5y​

Answer 1

Solution:

The Given Polynomial is

• 3y² -8 + 5y

Now, Arrange this Polynomial in Quadratic Polynomial i.e ax² + bx + c

let Polynomial be f(x) = 3y² +5y- 8

☛ 3y² +5y- 8

☛ 3y² - 3y + 8y - 8

☛ 3y(y - 1) + 8(y-1)

☛ (3y+8)(y-1)

So, Zeroes of Polynomial be;

f(x) = 3y+8= 0. 0r f(x) = y - 1

f(x)= 3y = -8. 0r f(x) = y - 1 => 0

☛ y = 8/3. 0r y = 1

Therefore, Zeroes of Polynomial are 8/3 and 1.