How do you factor #6r^2-28r+16#?

2 Answers
George C.
Jun 2, 2015

#6r^2-28r+16#

#=2(3r^2-14r+8)#

To factor #3r^2-14r+8# use a modified AC Method...

#A=3#, #B=14#, #C=8#

Look for a factorization of #AC=3xx8=24# into a pair of factors whose sum is #B=14#.

The pair #B1=2#, #B2=12# works.

Then for each of the combinations: #(A, B1)# and #(A, B2)#, divide by the HCF (highest common factor) to get the coefficients of a factor of #3r^2-14r+8# ...

#(3, 2)# (HCF #1#) #-> (3, 2) -> (3r-2)#
#(3, 12)# (HCF #3#) #-> (1, 4) -> (r - 4)#

So

#6r^2-28r+16 = 2(3r-2)(r-4)#

EZ as pi
Aug 26, 2016

#2(3x-2)(x-4)#

Explanation:

Factoring requires a solid knowledge of the multiplication tables.

Always look for a common factor first - all these numbers are even.

#6r^2-28r +16 =2(3r^2-14r +8) " factor the trinomial"#

All the clues are in the trinomial o3 3 and 8. Compare the colors and read it as follows:

#color(orange)(3) color(blue)(-) color(red)(14) color(lime)(+) color(orange)(8)#

Find the factors of #color(orange)(3and 8)# which #color(lime)("ADD")# to give #color(red)(14)#

The signs will be #color(lime)("THE SAME")#, they are both#color(blue)(" minus")#

Use different factors of 3 and 8 and cross-multiply with different combinations, until the sum of the products is 14.

#color(white)(xxxx)color(orange)((3)" "(8))#
#color(white)(xxxxx) 3" "2 rarr 1xx2 =2#
#color(white)(xxxxx) 1" "4 rarr 3xx4 =ul12#
#color(white)(xxxxxxxxxxxxxxxxxx)14 larr "we have the correct factors"#

Now put the signs into the brackets.

#2(3x-2)(x-4)#

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