How do you factor #a^2 - 28a + 192 = 0#?

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Answer 1 · 2018-05-30T19:21:55

See a solution process below:

We can factor this as:

#(a - 16)(a - 12) = 0#

To solve this we can equate each term on the left to #0# and solve for #a#:

Solution 1)

#a - 16 = 0#

#a - 16 + color(red)(16) = 0 + color(red)(16)#

#a - 0 = 16#

#a = 16#

Solution 2)

#a - 12 = 0#

#a - 12 + color(red)(12) = 0 + color(red)(12)#

#a - 0 = 12#

#a = 12#

The Solution Set Is:

#a = {12, 16}#