Question

solve the equation.
$y^3 + 3y^2 = 49y + 147$

the solution set is {}.
(simplify your answer. type an exact answer, using radicals as ne
solution only once.)

Explanation:

Step1: Rearrange the equation

Move all terms to the left side: \( y^3 + 3y^2 - 49y - 147 = 0 \)

Step2: Factor by grouping

Group the first two and last two terms: \( (y^3 + 3y^2) + (-49y - 147) = 0 \)
Factor out common terms: \( y^2(y + 3) - 49(y + 3) = 0 \)

Step3: Factor out \((y + 3)\)

We get: \( (y + 3)(y^2 - 49) = 0 \)

Step4: Factor \(y^2 - 49\)

Using difference of squares (\(a^2 - b^2=(a - b)(a + b)\)): \( (y + 3)(y - 7)(y + 7) = 0 \)

Step5: Solve for \(y\)

Set each factor equal to zero:

• \( y + 3 = 0 \Rightarrow y = -3 \)
• \( y - 7 = 0 \Rightarrow y = 7 \)
• \( y + 7 = 0 \Rightarrow y = -7 \)

Answer:

\(\{-7, -3, 7\}\)