**How to find complex roots of a 4th degree polynomial :**

Let us see some example problems to understand the above concept.

**Example 1 :**

Solve the equation x⁴ − 4x² + 8x + 35 = 0, if one of its roots is 2 + 3 i.

**Solution :**

Since the complex number 2 + i√3 is one root, then its conjugate 2 - i√3 is also a root.

Now we are going to form a quadratic equation with these two roots.

General form of a quadratic equation with roots a and b is

x² - (Sum of the roots) x + Product of the roots = 0

Sum of the roots = 2 + i√3 + 2 - 3i ==> 4

Product of roots = (2 + i√3) (2 - i√3) = 2² - (i√3)²

= 4 - 3 (-1) ==> 4 + 3 ==> 7

x² - 4 x + 7 = 0

Actually we have a polynomial of degree 4, we can split the given polynomial into two quadratic equations.

So far from the given roots we found a quadratic equation. From this quadratic we are going to find other part,

x⁴ − 4x² + 8x + 35 = (x² - 4 x + 7) (x² - P x + 5)

In order to find the value of p, we are equate the coefficients of x term

8 = -7p - 20

Add 20 on both sides

8 + 20 = -7p

-7p = 28

divide by -7 on both sides

p = 28/(-7) ==> -4

x² -(-4) x + 5

The other part of the given polynomial is x² + 4 x + 5. By solving this quadratic equation we will get two roots.

We cannot factor this quadratic equation. So we are going to use the quadratic formula to solve this equation.

a = 1, b = 4 and c = 5

x = -b ±√(b² -4ac)/2a

x = -4 ±√(4² -4(1)(5)/2(1)

x = -4 ±√(16-20)/2(1)

x = -4 ±√(-4)/2(1)

x = (-4 ± 2i)/2

x = -2 ± i

Hence the roots are 2 + i√3, 2 - i√3, -2 + i, -2 - i

Let us see another example of the topic "how to find complex roots of a 4th degree polynomial". how to find complex roots of a 4th degree polynomial

**Example 2 :**

Solve the equation x⁴ − 8x³ + 24x² - 32x + 20 = 0, if one of its roots is 3 + i.

**Solution :**

Since the complex number 3 + i is one root, then its conjugate 3 - i is also a root.

Now we are going to form a quadratic equation with these two roots.

General form of a quadratic equation with roots a and b is

x² - (Sum of the roots) x + Product of the roots = 0

Sum of the roots = 3 + i + 3 - i ==> 6

Product of roots = (3 + i) (3 - i) = 3² - i²

= 9 - (-1) ==> 9 + 1 ==> 10

x² - 6 x + 10 = 0

Actually we have a polynomial of degree 4, we can split the given polynomial into two quadratic equations.

So far from the given roots we found a quadratic equation. From this quadratic we are going to find other part,

x⁴ − 8x³ + 24x² - 32x + 20 = (x² - 6 x + 10) (x² - P x + 2)

In order to find the value of p, we are equate the coefficients of x term

-32 = -10p - 12

Add 12 on both sides

-32 + 12 = -10p

-10p = -20

divide by -10 on both sides

p = -20/(-10) ==> 2

x² -(2) x + 5

The other part of the given polynomial is x² - 2 x + 5. By solving this quadratic equation we will get two roots.

We cannot factor this quadratic equation. So we are going to use the quadratic formula to solve this equation.

a = 1, b = -2 and c = 5

x = -b ±√(b² -4ac)/2a

x = 2 ±√(2² -4(1)(5)/2(1)

x = 2 ±√(4-20)/2(1)

x = 2 ±√(-16)/2(1)

x = (2 ± 4i)/2

x = 1 ± 2i

Hence the roots are 3 + i, 3 - i, 1 + 2i, 1 - 2i

After having gone through the stuff given above, we hope that the students would have understood "how to find complex roots of a 4th degree polynomial".

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