How do you factor the trinomial #10y^2+22y-24#?

2 Answers
Feb 26, 2016

#2(5y-4)(y+3)#

Explanation:

First, recognize that there is a common factor of #2# in each term, which can be factored out of the expression.

#10y^2+22y-24=2(5y^2+11y-12)#

To factor the remaining trinomial #5y^2+11y-12#, we need to look for two numbers that meet the following criteria:

  • Their sum is the coefficient of the middle term #mathbf((11))#
  • Their product is the product of the coefficients of the first and last terms #mathbf("(")5xx-12=mathbf(-60")")#

We should examine the factor pairs of #-60# and try to find a pair whose sum is #11#.

#{:("-1 and 60",",","1 and -60"),("-2 and 30",",","2 and -30"),("-3 and 20",",","3 and -20"),(color(blue)"-4 and 15",",","4 and -15"),("-5 and 12",",","5 and -12"),("-6 and 10",",","6 and -10"):}#

We should use the pair #-4,15# to rewrite #11y#:

#2(5y^2+11y-12)=2(5y^2+15y-4y-12)#

Now, factor by grouping: factor a common terms from each smaller "set" of two in the term:

#2(color(red)(5y^2+15y)-color(green)(4y-12))=2[color(red)((5y^2+15y))-color(green)((4y+12))]#

Notice that the sign of the #12# changed since a negative was factored from the second term.

#2[(5y^2+15y)-(4y+12)]=2[5y(y+3)-4(y+3)]#

Factor a common #(y+3)# term from the #5y# and #-4# terms. Recall the #2# will stay "outside," just hanging around.

#2[5y(y+3)-4(y+3)]=ul(2(5y-4)(y+3)#

This is fully factored.

Feb 26, 2016

#f(y) = 2(5y - 4)(y + 3)#

Explanation:

#f(y) = 2f'(y) =2(5y^2 + 11y - 12)#
I use the new AC Method to factor trinomials (Socratic Search)
#f'(y) = 5y^2 + 11y - 12 = # 5(x + p)(x + q)
Converted trinomial #f''(y) = y^2 + 11y - 60 =# (y + p')(y + q').
p' and q' have opposite signs.
Compose factor pairs of (ac = -60) --> ...(-3, 20)(-4, 15). This sum is 11 = b. Then, p' = -4 and q' = 15.
Back to original trinomial f'(y): p = (p')/a = -4/5 and q = (q')/a = 15/5 = 3.
Factored form: #f(y) = 2(5)(y - 4/5)(y + 3) =# 2(5y - 4)(y + 3)