How do you factor #y= -12r^2+5r+3# ?

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Answer 1 · 2016-01-05T17:29:32

#(3r + 1 )(-4r + 3)# or indeed #(3r + 1)( 3 - 4r)#

to factor this trinomial # - 12r^2 + 5r + 3 #

compare to general form: # ax^2 + bx +c #

first multiply ac in this case # - 12 xx 3 = - 36 #

now consider the factors of - 36 which also add together to give b =5 in this question .

the pairs of factors of - 36 are ( - 2 , 18 ) reject as sum ≠ 5

( - 3 , 12 ) reject as sum ≠ 5

( - 4 , 9 ) accept as sum = 5

now rewrite the function as # -12r^2 - 4r + 9r + 3 #

(Note 5r has been replaced by #-4r + 9r#)

factor the expression in pairs to give: #-4r (3r + 1 ) + 3 ( 3r + 1 )#

#( 3r + 1 )# is a common factor here

#( 3r + 1)( - 4r + 3 )#3

# rArr - 12r^2 + 5r + 3 = ( 3r + 1 )(- 4r + 3 ) #