Question

Factor the given polynomial completely. 12y⁴-66y³+30y²

Answer 1

$12y {}^{4} - 66y {}^{3} + 30y {}^{2} \\ 6y {}^{2} (2y {}^{2} - 11y + 5) \\ 6y {}^{2} (2y {}^{2} - y - 10y + 5) \\ 6y {}^{2} (y(2y - 1) - 5(2y - 1)) \\ 6y {}^{2} (2y - 1)(y - 5)$

Answer 2

Answer:

6y²(2y - 1)(y - 5)

Step-by-step explanation:

12$y^{4}$ - 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$y^{4}$ - 66y³ + 30y² = 6y²(2y - 1)(y - 5)