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Explanation of irrational root theorem and imaginary root theorem

The Irrational Root Theorem

The Irrational Root Theorem says if a + \sqrt{b} is also a root of observed polynomial. In other words, irrational roots come in conjugate pairs.

Irrational Root Theorem

Example 1. Find the rational and irrational roots of the following polynomial equation.

x^3 + x^2 - 3x - 3 = 0

If this equation has imaginary roots, by the Imaginary Root Theorem,  must divide 5.

a^2 + b^2 \in \{ 1, 5\}

Now we have to think all the ways these numbers can be written as the sum of two squares of complex numbers.

First, for the number 1: 1 = 0 + 1 = 0 + (\pm 1) . This means that for this case a + bi \in \{\pm i \}.

Second, for the number 5: 5 = 1 + 4 and 5 = 4 + 1. For this case a + bi \in \{ \pm1\pm2i, \pm2i \pm1\}.

Now we have 9 possible solutions:

a + bi \in \{ \pm1\pm2i, \pm2i \pm1\}

To eliminate few of these solution we’ll use the noted addition to the Imaginary Root Theorem.

We’ll use k = - 1. (a - (- 1))^2 + b^2 must divide f(- 1).

a + 1 + b^2 must be divisible by 2. We’ll use this test for every possible solution. In the end we are left with:

a + bi \in \{ \pmi, - 2 \pm i \}

x_1 = - 2 + i, x_2 = - 2 - i

After finding these two solutions we have a quadratic equation left to solve. This will lead us to the remaining two solutions to the given equation:

x_3 = \frac{1}{2} (- 1 + i\sqrt{3} ), x_4 = \frac{1}{2} (- 1 - i\sqrt{3} )

Imaginary root theorem

As an addition to this theorem, for every whole number k, (a - k)^2 + b^2  divides f(k).

Example 2. Find all the roots of the following polynomial equation:

x^4 + 5x^3 + 10x^2 + 9x + 5 = 0

If this equation has imaginary roots, by the Imaginary Root Theorem, a^2 + b^2 must divide 5.

a^2 + b^2 \in \{ 1, 5 \}

Now we have to think all the ways these numbers can be written as the sum of two squares of complex numbers.

First, for the number 1: 1 = 0 + 1 = 0 + (\pm 1)^2 . This means that for this case a + bi \in \{ \pm 1 \}.

Second, for the number 5: 5 = 1 + 4 and 5 = 4 + 1. For this case a + bi \in \{ \pm 1 \pm 2i, \pm 2i \pm 1 \}

Now we have 9 possible solutions:

a + bi \in \{ \pm i, \pm 1 \pm 2i, \pm 2i \pm 1 \}

To eliminate few of these solution we’ll use the noted addition to the Imaginary Root Theorem.

We’ll use k = - 1. (a - (- 1))^2 + b^2 must divide f(- 1).

a + 1 + b^2 must be divisible by 2. We’ll use this test for every possible solution. In the end we are left with:

a + bi \in \{ \pm i, - 2 \pm i \}

x_1 = -2 + i, x_2 = - 2 - i

After finding these two solutions we have a quadratic equation left to solve. This will lead us to the remaining two solutions to the given equation:

x_3 = \frac{1}{2}(- 1 + i\sqrt{3}), x_4 = \frac{1}{2}(- 1 - i\sqrt{3})

 

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