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(Note 5r has been replaced by $-4r + 9r$ )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

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

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