How do you find the zeroes for #3x^4 - 4x^3 + 4x^2 - 4x + 1#?

2 Answers
Sep 15, 2015

Answer:

The zeros are #x=pm i, 1/3, 1#

Explanation:

The Rational Roots Theorem implies that if #f(x)=3x^4-4x^3+4x^2-4x+1# has any rational roots, they must be #pm 1# or #pm 1/3# (the numerator must divide the constant term and the denominator must divide the coefficient of the highest power).

If you check #1/3# and #1#, you'll see that they work (you can graph the function to guess that they work): #3*(1/3)^4-4*(1/3)^3+4*(1/3)^2-4*(1/3)+1=1/27-4/27+12/27-36/27+27/27=0# and #3-4+4-4+1=0#.

Doing long division (or synthetic division) leads to #f(x)=3x^4-4x^3+4x^2-4x+1=(x-1)(3x^3-x^2+3x-1)=(x-1)(3x-1)(x^2+1)=(x-1)(3x-1)(x-i)(x+i)#

Sep 15, 2015

Answer:

Use the rational zeros theorem (or the sum of the coefficients "shortcut") to find that #1# is a zero, divide by #x-1# and factor.

Explanation:

#3x^4 - 4x^3 + 4x^2 - 4x + 1 = 0# at #x=1# instead of test ing the other possible rational zeros, divide first:

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

So,

#3x^4 - 4x^3 + 4x^2 - 4x + 1 = (x-1)[underbrace(3x^3-x^2)+underbrace(3x-1)]#

# = (x-1)[x^2(3x-1)+1(3x-1)]#

# = (x-1)[(x^2+1)(3x-1)]#

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

The real zeros are #1# and #1/3# and imaginary zeros are #+-sqrt(-1)#