How do you factor #8v^2-30v+18#?

2 Answers
Apr 26, 2016

2(4v - 3)(v - 3)

Explanation:

Use the systematic new AC Method (Socratic Search).
#f(v) = 8v^2 - 30v + 18 =# 8(v + p)(v + q)
Converted trinomial: #f'(v) = v^2 - 30v + 144 =# (v + p')(v + q').
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 144) ---> (4, 36)(-4, -36)(6, 24)(-6, -24)..This sum is (-30 = b). Then p' = -6 and q' = -24.
Back to original f(v), #p = (p')/a = -6/8 = -3/4#, and #q = (q')/a = -24/8 = -3#.
Factored form: #f(v) = 8(v - 3/4)(v - 3) = 2(4v - 3)(v - 3).#

Apr 26, 2016

Solution reworked

Explanation:

Given;#" "8v^2-30v+18#

Sometimes you can spot the factors that work easily. Other times it takes a little work.

Notice that the coefficients and constant are even. So we can factor 2 out giving:

#" "2(4v^2-15v+9)#

So now we see if we can factorise #(4v^2-15+9)#
Consider the whole number factors of 4#->#{1,4} ; {2,2}
Consider the whole number factors of 9#->#{1,9} ; {3,3}
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Notice that #(3xx4)+(3xx1)=12+4=15#
But we need #-15# so we need to configure this multiplication to give as that value.

Try:#" "(4v-3)(v-3)# this gives:#" "4v^2-12v-3v+9# This works
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the full factorisation is:

#color(blue)(" "2(4v-3)(v-3))#

#color(red)("If this factorisation is correct it should have the same graph curve as the original equation."#

Tony B