Question

solve.
$y^{3}+4y - y^{2}-4 = 0$
$y = \square$
(simplify your answer, including any radicals and $i$ as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)

Explanation:

Step1: Rearrange and factor by grouping

Given the equation \( y^{3}+4y - y^{2}-4 = 0 \). Rearrange the terms: \( y^{3}-y^{2}+4y - 4 = 0 \). Group the first two terms and the last two terms: \( (y^{3}-y^{2})+(4y - 4)=0 \). Factor out the greatest common factor from each group: \( y^{2}(y - 1)+4(y - 1)=0 \).

Step2: Factor out the common binomial

Now, factor out the common binomial \( (y - 1) \) from the two terms: \( (y - 1)(y^{2}+4)=0 \).

Step3: Solve for y using the zero - product property

The zero - product property states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).

• Case 1: \( y-1=0 \), then \( y = 1 \).
• Case 2: \( y^{2}+4=0 \). Solve for \( y \): \( y^{2}=- 4 \), so \( y=\pm\sqrt{-4}=\pm2i \).

Answer:

\( y = 1, - 2i, 2i \)