Intermediate Algebra, Chapter 5, 5.2, Section 5.2, Problem 83

Solve $[-(y^4 - y^2 + 1) -(y^4 + 2y^2 + 1)] + (3y^4 - 3y^2 - 2)$
Remove the parentheses that are not needed from the expression.

$−(y^4−y^2+1)−(y^4+2y^2+1)+3y^4−3y^2−2$


Multiply $−1$ by each term inside the parentheses.

$−y^4+y^2−1−(y^4+2y^2+1)+3y^4−3y^2−2$


Multiply $−1$ by each term inside the parentheses.

$−y^4+y^2−1−y^4−2y^2−1+3y^4−3y^2−2$


Since $−y^4$ and $−y^4$ are like terms, subtract $y^4$ from $−y^4$ to get $−2y^4$.

$−2y^4+y^2−1−2y^2−1+3y^4−3y^2−2$


Since $−2y^4$ and $3y^4$ are like terms, subtract $3y^4$ from $−2y^4$ to get $y^4$.

$y^4+y^2−1−2y^2−1−3y^2−2$


Since $y^2$ and $−2y^2$ are like terms, add $−2y^2$ to $y^2$ to get $−y^2$.

$y^4−y^2−1−1−3y^2−2$


Since $−y^2$ and $−3y^2$ are like terms, subtract $3y^2$ from $−y^2$ to get $−4y^2$.

$y^4−4y^2−1−1−2$


Subtract 1 from $−1$ to get $−2$.

$y^4−4y^2−2−2$


Subtract 2 from $−2$ to get $−4$.

$y^4−4y^2−4$