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

Sep 15, 2015

The zeros are $x = \pm i , \frac{1}{3} , 1$

#### Explanation:

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

If you check $\frac{1}{3}$ and $1$, you'll see that they work (you can graph the function to guess that they work): $3 \cdot {\left(\frac{1}{3}\right)}^{4} - 4 \cdot {\left(\frac{1}{3}\right)}^{3} + 4 \cdot {\left(\frac{1}{3}\right)}^{2} - 4 \cdot \left(\frac{1}{3}\right) + 1 = \frac{1}{27} - \frac{4}{27} + \frac{12}{27} - \frac{36}{27} + \frac{27}{27} = 0$ and $3 - 4 + 4 - 4 + 1 = 0$.

Doing long division (or synthetic division) leads to $f \left(x\right) = 3 {x}^{4} - 4 {x}^{3} + 4 {x}^{2} - 4 x + 1 = \left(x - 1\right) \left(3 {x}^{3} - {x}^{2} + 3 x - 1\right) = \left(x - 1\right) \left(3 x - 1\right) \left({x}^{2} + 1\right) = \left(x - 1\right) \left(3 x - 1\right) \left(x - i\right) \left(x + i\right)$

Sep 15, 2015

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:

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

$\frac{3 {x}^{4} - 4 {x}^{3} + 4 {x}^{2} - 4 x + 1}{x - 1} = 3 {x}^{3} - {x}^{2} + 3 x - 1$

So,

$3 {x}^{4} - 4 {x}^{3} + 4 {x}^{2} - 4 x + 1 = \left(x - 1\right) \left[\underbrace{3 {x}^{3} - {x}^{2}} + \underbrace{3 x - 1}\right]$

$= \left(x - 1\right) \left[{x}^{2} \left(3 x - 1\right) + 1 \left(3 x - 1\right)\right]$

$= \left(x - 1\right) \left[\left({x}^{2} + 1\right) \left(3 x - 1\right)\right]$

$= \left(x - 1\right) \left(3 x - 1\right) \left({x}^{2} + 1\right)$

The real zeros are $1$ and $\frac{1}{3}$ and imaginary zeros are $\pm \sqrt{- 1}$