2y^2- 14y+ 20
By splitting the middle term:
2y^2 -10y -4y +20
By taking the common outside:
2y(y-5) -4(y-5)
=(2y-4)(y-5)
=2(y-2)(y-5)
Therefore D is the right answer
Answer:
D
Step-by-step explanation:
Given
2y² - 14y + 20 ← factor out 2 from each term
= 2(y² - 7y + 10)
To factor the quadratic
Consider the factors of the constant term (+ 10) which sum to give the coefficient of the y- term (- 7)
The factors are - 2 and - 5, since
- 2 × - 5 = 10 and - 2 - 5 = - 7, so
y² - 7y + 10 = (y - 2)(y - 5) and
2y² - 14y + 20 = 2(y - 2)(y - 5) → D
