Answer to Question #136009 in Math for Rufuson Oyinki

Question

d). Solve for x in the following polynomials
(i). f(x) = 3x4 + 2x3 – 15x2 + 12x – 6
(ii). 2x4 + 3x3 – 13x2 – 6x + 8 = 0

Answer 1

(i) 3x⁴+2x³-15x²+12x-6=0

Let f(x)=3x⁴+2x³-15x²+12x-6

d/dx(3x⁴+2x³-15x²+12x-6)=12x³+6x²-30x+12

∴f′(x)=12x³+6x²-30x+12

Here f(1) = - 4 < 0 and f (2) = 22 > 0

∴ Root lies between 1 and 2

x₀=1+22=1.5

x₀=1.5

1st iteration :

f(x₀)=f(1.5)=3⋅1.54+2⋅1.53-15⋅1.52+12⋅1.5-6=0.1875

f′(x₀)=f′(1.5)=12⋅1.53+6⋅1.52-30⋅1.5+12=21

x₁=x₀-f(x₀)f′(x₀)

x₁=1.5-0.187521

x₁=1.4911

2nd iteration :

f(x₁)=f(1.4911)=3⋅1.49114+2⋅1.49113-15⋅1.49112+12⋅1.4911-6=0.0027

f′(x₁)=f′(1.4911)=12⋅1.49113+6⋅1.49112-30⋅1.4911+12=20.3887

x₂=x₁-f(x₁)f′(x₁)

x₂=1.4911-0.002720.3887

x₂=1.4909

Approximate root of the equation 3x4+2x3-15x2+12x-6=0 using Newton Raphson method is 1.4909.

Now again using Newton Raphson Method to find the second root, now assuming initial condition to be -3 because while putting -3, the equation becomes nearer to zero. So the roots may be nearer to -3.

f(x)=3x⁴+2x³-15x²+12x-6

x₀ = - 3

1st iteration :

 

f(x₀)= f(-3) = 3⋅(-3)4+2⋅(-3)3-15⋅(-3)2+(12-3)-6=12

f′(x₀)= f′(-3) =

12⋅(-3)3+6⋅(-3)2-(30-3)+12=-168

x₁ = x₀ - f(x₀)f′(x₀)

x₁ = -3-12-168

x₁ = -2.9286

 

2nd iteration :

 

f(x₁) = f(-2.9286) = 3⋅(-2.9286)4+2⋅(-2.9286)3-15⋅(-2.9286)2+(12-2.9286)-6=0.6459

f′(x₁) = f′(-2.9286) = 12⋅(-2.9286)3+6⋅(-2.9286)2-(30-2.9286)+12=-150.0875

x₂ = x₁ - f(x1)f′(x1)

x₂ = -2.9286-0.6459-150.0875

x₂ = -2.9243

 

3rd iteration :

 

f(x₂) =  f(-2.9243) = 3⋅(-2.9243)4+2⋅(-2.9243)3-15⋅(-2.9243)2+(12-2.9243)-6=0.0023

f′(x₂) = f′(-2.9243) = 12⋅(-2.9243)3+6⋅(-2.9243)2-(30-2.9243)+12=-149.041

x₃ = x₂ - f(x₂)f′(x₂)

x₃ = -2.9243-0.0023-149.041

x₃ = -2.9243

4th iteration :

 

f(x₃) = f(-2.9243)=3⋅(-2.9243)4+2⋅(-2.9243)3-15⋅(-2.9243)2+(12-2.9243)-6=0

f′(x₃) = f′(-2.9243)=12⋅(-2.9243)3+6⋅(-2.9243)2-(30-2.9243)+12=-149.0373

x₄ = x₃ - f(x₃)f′(x₃)

x₄ = -2.9243-0-149.0373

x₄ = -2.9243

Approximate root of the equation 3x⁴+2x³-15x²+12x-6=0 using Newton Raphson method is -2.9243

So, The two roots of our polynomial equation are found to be - 2.9243 and 1.4909.

The remaining two roots are not complex so it can’t be calculated.

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(ii). 2x⁴ + 3x³ – 13x² – 6x + 8 = 0

By Hit and Trial method, Putting x = -1 in the above equation:

 

2(-1)⁴ + 3(-1)³ – 13(-1)² – 6(-1) + 8 = 0

0 = 0

 

Therefore, x = -1 is a solution of the above polynomial equation.

 

By hit and trial method, again putting x = 2 in the above equation:

 

2(2)⁴ + 3(2)³ – 13(2)² – 6(2) + 8 = 0

0 = 0

 

Therefore, x = 2 is also a solution of the above polynomial equation.

 

Now the two roots of this equations are found to be -1, 2.

 

On dividing (2x⁴ + 3x³ – 13x² – 6x + 8) by (x+1).(x-2), we get:

 

( 2x⁴ + 3x³ – 13x² – 6x + 8)/((x+1)(x-2)) = 2x² + 5x – 4

 

Now the equation can be written in the form,

 

2x⁴ + 3x³ – 13x² – 6x + 8 = 0

(x + 1) (x - 2) (2x² + 5x - 4) = 0

 

Using Quadratic formula on 2x² + 5x – 4, we get:

x = $-5/4±√(〖(5〗^2+4*2*4)/4$

x = $-5/4+√(57/4)$ and x = $-5/4-√(57/4)$

So, the all roots of the equation are found to be

Solution: -1, 2, -5/4+√(57/4)  and -5/4-√(57/4)