How do you factor #4x ^ { 2} - 16x = 64#?

1 Answer
Nghi N
Jul 2, 2017

#f(x) = 4(x - 2 - sqrt5)(x - 2 + sqrt5)#

Explanation:

#f(x) = 4y = 4(x^2 - 4x - 16)#
To factor #y = x^2 - 4x - 16#, find the 2 real roots of y = 0.
#D = d^2 = b^2 - 4ac = 16 + 64 = 80# --> #d = +- 4sqrt5#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 4/2 +- 4sqrt5/2 = #2 +- sqrt5# Factored form of y: #y = a( x - x1)(x - x2) = (x - 2 - sqrt5)(x - 2 + sqrt5)# #f(x) = 4y = 4(x - 2 - sqrt5)(x - 2 + srqt5)#

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