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

Question

Hazel Anglin · Accepted Answer

#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)#

Hazel Anglin · Answer

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

A strong understanding of multiplication tables is necessary for factoring.

The first thing to look for is a common factor: all of these numbers are even.

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

The trinomial o3 3 and 8 contains all of the clues. 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")#

Utilize various factors of 3 and 8 and perform cross-multiplication using various combinations until the total of the products reaches 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"#

Place the signs inside the brackets now.

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

Savannah Anderson · Answer

undefined To factor the quadratic expression $6r^2 - 28r + 16$, first, identify any common factors among the coefficients. In this case, the greatest common factor (GCF) is 2. Next, use the quadratic formula or factoring techniques to factor the expression. To factor $6r^2 - 28r + 16$, you can first factor out the GCF, which is 2: $2(3r^2 - 14r + 8)$. Now, focus on factoring the trinomial inside the parentheses. You can do this by finding two numbers that multiply to the product of the coefficient of the $r^2$ term (3) and the constant term (8), which is 24, and add up to the coefficient of the $r$ term (-14). The numbers are -6 and -4. Rewrite the middle term (-14r) using these two numbers: $2(3r^2 - 6r - 8r + 8)$. Factor by grouping: $2(3r(r - 2) - 8(r - 2))$. Factor out the common binomial factor $r - 2$: $2(r - 2)(3r - 8)$. So, the factored form of $6r^2 - 28r + 16$ is $2(r - 2)(3r - 8)$.