Question

find \\((2y^2 - 6y + 3)(4y^2 - y + 1)\\).

Explanation:

Step1: Distribute \(2y^2\)

\(2y^2 \times 4y^2 = 8y^4\), \(2y^2 \times (-y) = -2y^3\), \(2y^2 \times 1 = 2y^2\)

Step2: Distribute \(-6y\)

\(-6y \times 4y^2 = -24y^3\), \(-6y \times (-y) = 6y^2\), \(-6y \times 1 = -6y\)

Step3: Distribute \(3\)

\(3 \times 4y^2 = 12y^2\), \(3 \times (-y) = -3y\), \(3 \times 1 = 3\)

Step4: Combine like terms

• \(y^4\) term: \(8y^4\)
• \(y^3\) terms: \(-2y^3 -24y^3 = -26y^3\)
• \(y^2\) terms: \(2y^2 + 6y^2 + 12y^2 = 20y^2\)
• \(y\) terms: \(-6y -3y = -9y\)
• Constant term: \(3\)

So the result is \(8y^4 -26y^3 + 20y^2 -9y + 3\)

Answer:

\(8y^4 - 26y^3 + 20y^2 - 9y + 3\)