4(y-1)^2-2(y+3)^2-(2y+1)^2, please explain in steps how to answer this question, it is driving me insane!!!!!!

I keep getting the same wrong answer, the answer is supposed to be -2y^2-24y-15

thank you

Calm down, take a deep breath, and work through it a step at a time. First expand the squares, then multiply through by the coefficients, and then it becomes simple to just group the corresponding terms (e.g., y^2's together, y's together, and constants together).

4(y-1)^2 - 2(y+3)^2 - (2y+1)^2.
Expanding the squares, you get:
4(y^2 - 2y + 1) - 2(y^2 + 6y + 9) - (4y^2 + 4y + 1)
Multiplying through by the coefficients, you get:
4y^2 - 8y + 4 - 2y^2 - 12y - 18 - 4y^2 - 4y - 1
If you find it easier, group the y^2 terms together, the y terms together, and the constants together:
4y^2 - 2y^2 - 4y^2 - 8y - 12y - 4y + 4 - 18 - 1
Now just combine them to get:
-2y^2 - 24y - 15

Answer: -2y^2 - 24y - 15

I assume you're meant to be simplifying this. All that is required is that you expand the brackets in each case, and add/subtract like terms (like terms being the ones where there is the same exponent on the y)

:. 1) expand: 4(y^2 - 2y +1) - 2(y^2 + 6y + 9) - (4y^2 + 4y + 1)
2) Multiply outside into brackets: 4y^2 - 8y + 4 - 2y^2 - 12y - 18 - 4^2 - 4y - 1
3) collect and add like terms: (4 - 2 - 4)*y^2 + (-8 - 12 - 4)*y + (4 - 18 - 1) = -2y^2 - 24y - 15

Hopefully the working makes sense.

4(y - 1)^2 - 2(y + 3)^2 - (2y + 1)^2

squaring the binomials yields:

4(y^2 - 2y + 1) - 2(y^2 + 6y + 9) - (4y^2 + 4y + 1)

distributing the coefficients yields:

4y^2 - 8y + 4 - 2y^2 - 12y - 18 - 4y^2 - 4y - 1

combining like terms yields:

(4 - 2 - 4)y^2 + (-8 - 12 - 4)y + (4 - 18 - 1)

simplifying:

-2y^2 - 24y - 15

4(y - 1)² - 2(y + 3)² - (2y + 1)²

= 4(y² - 2y + 1) - 2(y² + 6y + 9) - (4y² + 4y + 1)

= 4y² - 8y + 4 - 2y² - 12y - 18 - 4y² - 4y - 1

= - 2y² - 24y - 15