is 3(2y^2-5)(y+1) completely factored
How many real number solutions exist for 2x2 + 8x + 8 = 0?
To determine if the expression 3(2y^2-5)(y+1) is completely factored, we need to check if any further simplification is possible.
First, let's simplify the expression inside the first set of parentheses: 2y^2 - 5.
This expression cannot be factored further because it is in quadratic form and cannot be written as a product of two binomials with rational coefficients.
Next, let's simplify the expression inside the second set of parentheses: y + 1.
This expression also cannot be factored further because it is already in its simplest form, a binomial.
Now, let's put everything back together:
3(2y^2 - 5)(y + 1)
Since we have checked and found that both expressions inside the parentheses cannot be factored further, we can conclude that the expression 3(2y^2-5)(y+1) is completely factored.