HOW TO FIND A CUBIC POLYNOMIAL WHOSE ZEROS ARE GIVEN

Find all cubic polynomials with zeros of :

Problem 1 :

±2, 3

Solution :

The zeroes of required polynomials are -2, 2 and 3. That is,

x = -2, x = 2 and x = 3

Converting them as factors.

(x + 2) (x - 2) and (x - 3)

By multiplying the factor.

 P (x) = a[(x² - 2x + 2x - 4) (x - 3)]

= a[(x² - 4) (x - 3)]

= a[x³ - 3x² - 4x + 12]

= a[x²(x - 3) - 4(x - 3)]

P(x) = a[(x² - 4) (x - 3)] where a ≠ 0

Hence the required polynomial is a(x² - 4) (x - 3).

Problem 2 :

-2, ± i

Solution :

The zeroes of required polynomials are -2, 2 and 3.

That is,

x = -2, x = i and x = -i

Converting them as factors.

(x + 2) (x - i) and (x + i)

By multiplying the factor.

P(x) = a[(x² - xi + 2x - 2i) (x + i)]

= a[x³ + ix² - x²i - xi² + 2x² + 2xi  - 2ix - 2i²]

= a[x³ + 2x² + x + 2]

= a[x²(x + 2) + 1(x + 2)]

= a[(x² + 1) (x + 2)] where a ≠ 0

Hence the required polynomial is a(x² - 4) (x - 3).

Problem 3 :

3, -1 ± i

Solution :

The zeroes of required polynomials are 3, -1 + i and -1 - i. That is,

x = 3, x = -1 + i and x = -1 - i

Finding the quadratic factor with -1 + i and -1 - i

Sum of roots = -1 + i - 1 - i ==>  -2

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

= 1² - i² ==> 2

Quadratic factor :

y = x² - 2x + 2

Linear factor :

(x - 3)

= a(x² + 2x + 2) where a ≠ 0

Hence the required polynomial is a(x - 3) (x² + 2x + 2).

Problem 4 :

-1, -2 ± √2

Solution :

The zeroes of required polynomials are -1, -2 + √2  and -2 - √2. That is,

x = -1, x = -2 + √2 and x = -2 - √2

Sum of the roots = (-2 + √2 - 2 - √2)

= -4

Products of the roots = (-2 + √2) (-2 - √2)

= 4 + 2√2 - 2√2 - 2

= 2

P(x) = a[x² - (sum of the roots)x + (products of the roots)]

= a(x² + 4x + 2) where a ≠ 0

Hence the required polynomial is a(x + 1) (x² + 4x + 2).