HOW TO FIND QUARTIC POLYNOMIAL FROM ZEROS

Find all quartic polynomials with zeros of :

Problem 1 :

±1, ±√2

Solution :

±1, ±√2

Let f(x) be the quadratic polynomial whose roots are +1 and -1.

Sum of roots = 1 + (-1)

= 1 - 1

= 0

Products of roots = (1) ⋅ (-1)

= -1

f(x) = x² - (sum of roots)x + products of roots

= x² - 0x + (-1)

f(x) = x² - 1 ---(1)

Let g(x) be the quadratic polynomial whose roots are +√2 and -√2.

Sum of roots = √2 + (-√2)

= √2 - √2

= 0

Products of roots = (√2) ⋅ (-√2)

= -2

g(x) = x² - (sum of roots)x + products of roots

= x² - 0x + (-2)

g(x) = x² - 2 ---(2)

From (1) and (2)

P(x) = a[f(x) ⋅ g(x)]

= a[(x² - 1) ⋅ (x² - 2)] where a ≠ 0

Problem 2 :

2, -1, ±i√3

Solution :

2, -1, ±i√3

x = 2

f(x) = (x - 2) ---(1)

x = -1

g(x) = (x  + 1) ---(2)

Let z(x) be the quadratic polynomial whose roots are +i√3 and -i√3.

Sum of roots = i√3 + (-i√3)

= i√3 - i√3

= 0

Products of roots = (i√3) ⋅ (-i√3)

= 3

z(x) = x² - (sum of roots)x + products of roots

= x² - 0x + 3

z(x) = x² + 3 ---(3)

From (1), (2) and (3)

P(x) = a[f(x) ⋅ g(x) ⋅ z(x)]

P(x) = a[(x - 2) (x + 1) (x² + 3)] where a ≠ 0

Problem 3 :

±√3, 1 ± i 

Solution :

±√3, 1 ± i

Let f(x) be the quadratic polynomial whose roots are +√3 and -√3.

Sum of roots = √3 + (-√3)

= √3 - √3

= 0

Products of roots = (√3) ⋅ (-√3)

= -3

f(x) = x² - (sum of roots)x + products of roots

= x² - 0x + (-3)

f(x) = x² - 3 ---(1)

Let g(x) be the quadratic polynomial whose roots are 1+ i and 1 - i.

Sum of roots = (1 + i) + (1 - i)

= 1 + i + 1 - i

= 2

Products of roots = (1 + i) ⋅ (1 - i)

= 1 - i + i - i²

= 2

g(x) = x² - (sum of roots)x + products of roots

= x² - 2x + 2 ---(2)

From (1) and (2)

P(x) = a[f(x) ⋅ g(x)]

= a[(x² - 3) ⋅ (x² - 2x + 2)] where a ≠ 0

Problem 4 :

2 ± √5, -2 ± 3i

Solution :

2 ± √5, -2 ± 3i

Let f(x) be the quadratic polynomial whose roots are 2 + √5 and 2 - √5.

Sum of roots = (2 + √5) + (2 - √5)

= 2 + √5 + 2 - √5

= 4

Products of roots = (2 + √5) ⋅ (2 - √5)

= 4 - 2√5 + 2√5 - 5

= -1

f(x) = x² - (sum of roots)x + products of roots

= x² - 4x + (-1)

f(x) = x² - 4x - 1 ---(1)

Let g(x) be the quadratic polynomial whose roots are -2 + 3i and -2  - 3i.

Sum of roots = (-2 + 3i) + (-2 - 3i)

= -2 + 3i - 2 - 3i

= -4

Products of roots = (-2 + 3i) ⋅ (-2 - 3i)

= 4 + 6i -6i + 9

= 13

g(x) = x² - (sum of roots)x + products of roots

= x² + 4x + 13 ---(2)

From (1) and (2)

P(x) = a[f(x) ⋅ g(x)]

= a[(x² - 4x - 1) ⋅ (x² + 4x + 13)] where a ≠ 0