How do you factor completely #14a^2 - 68a + 48#?

1 Answer

Answer:

#14a^2-68a+48=2(7a-6)(a-4)#

Explanation:

We have:

#14a^2-68a+48#

We can first factor out the Largest Common Factor of 14, 68, and 48:

  • #14 = 2xx7#
  • #68 = 2xx2xx17#
  • #48= 2xx2xx2xx2xx3#

The Largest Common Factor is 2:

#2(7a^2-34a+24)#

Now we look for factors in the form of #(ax+b)(cx+d)# where

  • #ac=7#
  • #bd=24#
  • #ad+bc=-34#

We know the factors for #7=7xx1#, so let's set #a=7, c=1#

The factors for #24=-1xx-24, -2xx-12, -3xx-8, -4xx-6# (because #ad+bc# is a negative number and we've set #ac# to be positive, we need these factors to be negative).

Let's try some factors and trial and error our way into this:

#acolor(white)(00000)bcolor(white)(0000)c color(white)(00000)dcolor(white)(000000)adcolor(white)(00000)bc color(white)(000)ad+bc#

#7color(white)(000)-8color(white)(000)1 color(white)(000)-3color(white)(000)-21color(white)(000)-8 color(white)(000)-29#

#7color(white)(000)-6color(white)(000)1 color(white)(000)-4color(white)(000)-28color(white)(000)-6 color(white)(000)-34#

This gives us:

#2(7a-6)(a-4)#