How do you find the zeroes for 3x^4 - 4x^3 + 4x^2 - 4x + 13x4−4x3+4x2−4x+1?

2 Answers
Bill K.
Sep 15, 2015

The zeros are x=pm i, 1/3, 1x=±i,13,1

Explanation:

The Rational Roots Theorem implies that if f(x)=3x^4-4x^3+4x^2-4x+1f(x)=3x4−4x3+4x2−4x+1 has any rational roots, they must be pm 1±1 or pm 1/3±13 (the numerator must divide the constant term and the denominator must divide the coefficient of the highest power).

If you check 1/313 and 11, 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=03⋅(13)4−4⋅(13)3+4⋅(13)2−4⋅(13)+1=127−427+1227−3627+2727=0 and 3-4+4-4+1=03−4+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)f(x)=3x4−4x3+4x2−4x+1=(x−1)(3x3−x2+3x−1)=(x−1)(3x−1)(x2+1)=(x−1)(3x−1)(x−i)(x+i)

Jim H
Sep 15, 2015

Use the rational zeros theorem (or the sum of the coefficients "shortcut") to find that 11 is a zero, divide by x-1x−1 and factor.

Explanation:

3x^4 - 4x^3 + 4x^2 - 4x + 1 = 03x4−4x3+4x2−4x+1=0 at x=1x=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-13x4−4x3+4x2−4x+1x−1=3x3−x2+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)

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