Answer:
6y²(2y - 1)(y - 5)
Step-by-step explanation:
12[tex]y^{4}[/tex] - 66y³ + 30y² ← factor out 6y² from each term
= 6y²(2y² - 11y + 5) ← factor the quadratic
consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term, that is
product = 2 × 5 = 10 and sum = - 11
the factors are - 1 and - 10
use these factors to split the y- term
2y² - y - 10y + 5 ( factor the first/second and third/fourth terms )
y(2y - 1) - 5(2y - 1) ← factor out (2y - 1) from each term
(2y - 1)(y - 5)
then
12[tex]y^{4}[/tex] - 66y³ + 30y² = 6y²(2y - 1)(y - 5)
[tex]12y {}^{4} - 66y {}^{3} + 30y {}^{2} \\ 6y {}^{2} (2y {}^{2} - 11y + 5) \\ 6y {}^{2} (2y {}^{2} - y - 10y + 5) \\ 6y {}^{2} (y(2y - 1) - 5(2y - 1)) \\ 6y {}^{2} (2y - 1)(y - 5)[/tex]
