How to solve-The equation

Question

2x^3 − x^2 + 2x + 12 = 0 has one real root and two complex roots. Showing your working

 

Answer 1

Identify Rational Zeros :

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

The equation is 2x³ - x² + 2x + 12 = 0.

If p/q is a rational zero, then p is a factor of 12 and q is a factor of 2.

The possible values of p are   ± 1, ± 3, and ± 4.

The possible values for q are ± 1 and  ± 2.

So, p/q = ± 1/2, ± 3/2, ± 1, ± 2, ± 3 and ± 4.

Make a table for the synthetic division and test possible real zeros.

p/q	2	- 1	2	12
- 1/2	2	- 2	3	21/2
1/2	2	0	2	11
- 3/2	2	- 4	8	0

Since f(- 3/2) = 0,  x = - 3/2 is a zero. The depressed polynomial is  2x² – 4x + 8 = 0.

Divide the depressed polynomial by 2.

x² – 2x + 4 = 0.

Since the depressed polynomial of this zero, x² – 2x + 4, is a quadratic, use the quadratic formula to find the roots of the related quadratic equation.

Solution of the quadratic equation : x = [- b ± √(b² - 4ac)]/2a.

Compare the given equation with standard form of the quadratic equation ax² + bx + c = 0.

a = 1, b = - 2 and c = 4.

Solution : x = [- (- 2) ± √((- 2)² - (4*1*4))]/2*1

= [2 ± √(4 - 16)]/2

= [2 ± √- 12]/2

= [2 ± 2√(- 3)]/2

= 1 ± √- 3

Substitute - 1 = i².

=[1 ± √(3i²)

⇒ x = 1 ± i√3.

The equation has one real zero at x = - 3/2 and two imaginary zeros at x = 1 - i√3 and x = 1 + i√3.