# FIND ALL COMPLEX AND REAL ROOTS OF HIGHER DEGREE POLYNOMIALS GIVEN ONE ROOT

## About "Find all complex and real roots of higher degree polynomials given one root"

Find all complex and real roots of higher degree polynomials given one root :

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

Example 1 :

Solve the equation x − 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,

x − 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

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 "Find all complex and real roots of higher degree polynomials given one root".  Find all complex and real roots of higher degree polynomials given one root

Example 2 :

Solve the equation x − 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,

x − 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

-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 "Find all complex and real roots of higher degree polynomials given one root".

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