How do you factor #90y ^ { 2} - 21y - 45#?

2 Answers
Jul 21, 2017

#(15y+9)(6y-5)#

Explanation:

#90y^2-21y-45#

Factorise.

#90y^2-75y+54y-45#

#15y(6y-5)+9(6y-5)#

#(15y+9)(6y-5)#

Jul 21, 2017

3(5y + 3)(6y - 5)

Explanation:

Use the new AC Method to factor trinomial (Socratic, Google Search).
f() = 90y^2 - 21y - 45 = 3Z #. Factor Z = (30x^2 - 7y - 15) = 30(y + p)(y + q) Converted trinomial: Z' = y^2 - 7y - 450 = (y + p')(y + q') Proceeding: Find p' and q', then, divide them by a = 30 Find 2 number p' and q', that have opposite signs, knowing the sum (- b = 7) and product (ac = -450). Use calculator to compose factor pairs of (-450) --> ...(15, -30)(18 - 25). This last sum is (-7 = b). Therefore, p' = 18 and q' = -25. Back to original Z # = (p')/a = 18/30 = 3/5# and #q= (q')/a = -25/30 = - 5/6#
Factored form:
Z = 30(y + 3/5)(y - 5/6) = (5y + 3)(6y - 5)
Finally,
f(y) = 3(5y + 3)(6y - 5)