(y+1)(y+3)(5y-2)=0
That is an equation. What is the question?
Are you supposed to multiply it out or say what the solution values of y are?
The equation you provided is a polynomial equation, specifically a quadratic equation, in the form (y+1)(y+3)(5y-2) = 0. The question can vary depending on what you are trying to find.
If you are asked to multiply it out, it means you need to expand the equation to simplify it into a polynomial expression. To do this, you can apply the distributive property to multiply each term inside the parentheses.
If you are asked to find the solution values of y, it means you need to solve the equation for y. To find the solution values, you can set each factor equal to zero and solve for y.
For example:
1. Setting y+1 = 0, we have y = -1 as one solution.
2. Setting y+3 = 0, we have y = -3 as another solution.
3. Setting 5y-2 = 0, we have y = 2/5 as the third solution.
Therefore, the solution values for y in the given equation are y = -1, y = -3, and y = 2/5.