How do you factor #3x^2-13x+4#?

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Answer 1 · 2015-05-19T01:30:48

You can find its roots and then turn them into factors.

Using Bhaskara to find the roots:

#(13+-sqrt(169-4(3)(4)))/6#
#(13+-11)/6#
#x_1=4#, which is the same as the factor #x-4=0#
#x_2=1/3#, which is the same as the factor#3x-1=0#

Factoring your quadratic function, then:

#3x^2-13x+4=(x-4)(3x-1)#

Answer 2 · 2015-05-19T01:46:22

There is another way. I use the new AC Method (Google, Yahoo Search) to factor trinomials

y = 3x^2 - 13x + 4 = (x - p)( - q)

Converted function: f'(x) = x^2 - 13x + 12.= (x - p')(x -q')

Compose factor pair of 12: (1, 12). This sum is 12 + 1 = 13 = -b.

Then p' = -1 and q' = -12.

Consequently, p = (p')/a = -1/3 and q' = (q')/a = -12/3 = -4

Factored form of y: y = (x - 1/3)(x - 4) = (3x - 1)(x - 4).