How do you factor and solve #16x^2-40x+25=0#?

2 Answers
Apr 9, 2018

Answer:

#x = -5/4#

Explanation:

To factor, we first look to see if there is an immediate factor that is common to all terms. In this case, there is not.

Next, we look to find a pair of numbers that when multiplied will give #25# and when added will give #-40#.

The #x^2# term has a non-one coefficient, so this will affect possible pairs.

The factors of #25# are:

#(1, 25)#
#(5,5)#

The factors of #16# are:

#(1,16)#
#(4,4)#

We need to select one pair from the top list and one pair from the bottom list. It is usually smart to start with the pairs that have two numbers close in value. So let's try #(5,5)# with #(4,4)#.

#(4x+5)(4x+5) = 16x^2 + 20x + 20x + 25 #

#= 16x^2+40x+25#

This one worked!

Hence, the factored form is:

#(4x+5)(4x+5) = (4x+5)^2#

Solving:

#(4x+5)^2 = 0#

#4x+5 = 0#

#4x = -5#

#x = -5/4#

Apr 9, 2018

Answer:

Please look below.

Explanation:

#16x^2 - 40x + 25 = 0#

Using the quadratic formula #x = (-b +-sqrt(b^2-4ac))/(2a)# and the values #a = 16# , #b =-40# and #c = 25# shows that

#x = (40 +-sqrt((-40)^2-4xx16xx25))/(2xx16)#
#x = (40 +- sqrt(1600 - 1600))/32#
#x = 40/32#
#x = 5/4#

Hence:
#16x^2 - 40x + 25 = 16(x-5/4)^2#
# = (4x-5)^2#