# `x^5 - 3x^4 + x^3 - x^2 - x + 6 = 0` Use Newton's method to find all roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

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`x^5-3x^4+x^3-x^2-x+6=0`

To solve this, using Newton's method, apply the formula:

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`x_(n+1)=x_n- (f(x_n))/(f'(x_n))`

Let the function be:

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`f(x) =x^5-3x^4+x^3-x^2-x+6`

Then, take its derivative.

`f'(x) =5x^4-12x^3+3x^2-2x-1`

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Then, plug-in f(x) and f'(x) to the formula.

`x_(n+1) = x_n - (x_n^5-3x_n^4+x_n^3-x_n^2-x_n+6)/(5x_n^4-12x_n^3+3x_n^2-2x_n-1)`

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To get the initial value of x, refer to the graph of the function. (See attached figure.)

Notice that curve has three x values when f(x)=0. These are near x=-1, x=1 and x=3. These will be our initial values for x.

Let's consider solve for the first zero.

`x_1=-1`

`x_(2) = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1) =-1.047619047`

`x_3= x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1) =-1.044517237`

`x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1)=-1.044503071`

`x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1)=-1.044503071`

Now, we have two approximations that have same eight decimal places. Thus, one of the solution is approximately **x=-1.04450307** .

Next, let's solve for the second zero.

`x_1 = 1`

`x_2 = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1)=1.428571428`

`x_3 = x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1)=1.336197712`

`x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1) =1.332589426`

`x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1)=1.332583155`

`x_6 = x_5 - (x_5^5-3x_5^4+x_5^3-x_5^2-x_5+6)/(5x_5^4-12x_5^3+3x_5^2-2x_4-1)=1.332583155`

Now that we have two approximations that agree with eight decimal places, then one of the solution is approximately **x=1.332583155**.

Next, let's solve for the third zero.

`x_1 = 3`

`x_2 = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1) =2.792079207`

`x_3 = x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1)=2.715701083`

`x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1) =2.705675033`

`x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1) =2.705512135`

`x_6 = x_5 - (x_5^5-3x_5^4+x_5^3-x_5^2-x_5+6)/(5x_5^4-12x_5^3+3x_5^2-2x_4-1)=2.705512093`

`x_7 = x_6 - (x_6^5-3x_6^4+x_6^3-x_6^2-x_6+6)/(5x_6^4-12x_6^3+3x_6^2-2x_6-1)=2.705512093`

Now, we have two approximations that have same eight decimal places. Thus, one of the solution is approximately **x=2.705512093**.

**Therefore, the approximate solution of`x^5-3x^4+x^3-x^2-x+6=0` are **

**`x={-1.04450307 ,1.332583155,2.705512093}.` **