How do you factor #4x^4-19x^3+16x^2-19x+12#?

1 Answer
Oct 1, 2015

Answer:

Use the rational root theorem, divide by the first factor found, then factor by grouping to find:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x^2+1)#

Explanation:

Let #f(x) = 4x^4-19x^3+16x^2-19x+12#

By the rational root theorem, any rational roots of #f(x) = 0# will be expressible in lowest terms as #p/q#, where #p, q in ZZ#, #q != 0#, #p# a divisor of the constant term #12# and #q# a divisor of the coefficient #4# of the leading term.

Also, since the signs of the terms are alternating, all the Real roots of #f(x) = 0# are positive.

That means that the only possible rational roots are:

#1/4#, #1/2#, #3/4#, #1#, #3/2#, #2#, #3#, #4#, #6#, #12#

Let's try a few:

#f(1/4) = 4/256-19/64+16/16-19/4+12 = (1-19+64-304+768)/64 = 446/64#

#f(1/2) = 4/16-19/8+16/4-19/2+12 = (2-19+32-76+96)/8 = 35/8#

#f(3/4) = 81/64 - 513/64 + 9 - 57/4 + 12 = (81-513+576-912+768)/64 = 0#

So #x = 3/4# is a root and #(4x-3)# is a factor.

Divide by #(4x-3)# to find:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x^3-4x^2+x-4)#

The remaining cubic factor factors nicely by grouping:

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

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

So putting this together, we have:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x^2+1)#

The remaining quadratic factor #x^2 + 1# has no simpler linear factors with Real coefficients since #x^2 + 1 >= 1 > 0# for all #x in RR#.

If you really want to factor it further:

#x^2+1 = (x-i)(x+i)#

where #i# is the imaginary unit.

Giving us:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x-i)(x+i)#