How do you factor #10y^2 - 29y + 10#?

2 Answers
Aug 29, 2015

Answer:

#10y^2-29y+10 = (5y-2)(2y-5)#

Explanation:

The matching coefficient of the #y^2# term and the constant term indicate the symmetry we see in this factorisation, essentially leaving two possibilities with integer coefficients to try:

#(10y-1)(y-10) = 10y^2-101y+10#

and

#(5y-2)(2y-5) = 10y^2-29y+10#

Aug 30, 2015

Answer:

Factor: #f(y) = 10y^2 - 29y + 10#

Ans: (5y - 2)(2y - 5)

Explanation:

#f(y) = 10y^2 - 29y + 10 =# 10(x + p)(x + q)
I use the new AC Method (Socratic search)
Converted trinomial: #f'(y) = y^2 - 29 y + 100 =# (y + p')(y + q')
p' and q' have same sign. Factor pairs of 100 --> (2, 50)(4, 25). This sum is 29 = -b. Then p' = -4 ans q' = -25.
Back to original trinomial: #p = -4/10 = -2/5# and #q = -25/10 = -5/2#.

Factoring form: #f(y) = (10)(y - 2/5)(y -5/2) = (5y - 2)(2y - 5)#