How do you factor #5x^2-22x+8#?

2 Answers
Aug 9, 2015

Answer:

Factor: y = 5x^2 - 22x + 8

Ans: (5x - 2)(x - 4)

Explanation:

Use the new AC Method to factor trinomial (Socratic Search):
#y = 5x^2 - 22x + 8 =# 5(x + p)(x + q)
Converted trinomial:
#y' = x^2 - 22x + 40 =# (x + p')(x + q')
Find p' and q', that have same sign, knowing sum (b = -22) and product (ac = 40). They are: p' = - 2 and q' = - 20
Therefor: #p = (p')/a = - 2/5# and #q = (q')/a = - 20/5 = - 4#
Factored form of y:
#y = 5(x - 2/5)(x - 4) = (5x - 2)(x - 4)#

Sep 11, 2017

Answer:

#(x-4)(5x-2)#

Explanation:

#5x^2 color(blue)(-22)xcolor(red)(+)8#

Find factors of #5 and 8# whose products #color(red)("ADD")# to #color(blue)(22)#

#5# is a prime number, so the factors are only #1 and 5 #

The factors of #8# are # 1 xx 8# or the other pair # 2 xx 4#

It may take a bit of trial and error, but notice that #5 xx 4 =20# which is close to the #20# we want.

#" "5 and 8" "# write the factors and cross multiply
#" "darr" "darr#
#" "1color(white)(xxxx)4" "rarr 5 xx 4 =20#
#" "5color(white)(xxxx)2" "rarr 1 xx 2 = ul2#
#color(white)(xxxxxxxxxxxxxxxxxxx)color(blue)(22)" "larr# the combination is correct.

The positive sign of #color(red)(+)8# indicates the following:

ADD the factors and the signs will be the SAME

The negative sign of #color(blue)(-22)# shows what the signs are.

The factors are #(1x-4)(5x-2)#