How do you factor #4x^2 - 5x + 1 = 0 #?

1 Answer
Feb 20, 2017

Answer:

#(4x-1)(x-1) = 0#

Explanation:

Given:

#4x^2-5x+1 = 0#

#color(white)()#
Method 1

Note that the sum of the coefficients is #0#. That is:

#4-5+1 = 0#

Hence #x=1# is a zero and #(x-1)# a factor:

#0 = 4x^2-5x+1 = (x-1)(4x-1)#

#color(white)()#
Method 2

Note that the prime factorisation of #451# is:

#451 = 11*41#

and this multiplication involves no carrying of digits.

Hence we find:

#4x^2+5x+1 = (x+1)(4x+1)#

and:

#4x^2-5x+1 = (x-1)(4x-1)#

#color(white)()#
Method 3

Note that:

#(ax-1)(bx-1) = abx^2-(a+b)x+1#

Since the constant term of the given example is #1#, look for a pair of factors of the leading coefficient #4# with sum matching the middle coefficient #5#. The pair #4, 1# works.

Hence we find:

#(4x-1)(x-1) = 4x^2-5x+1#