Newton's method. It works perfectly
There will be 3 solutions since the degree is 3.
Let initial x = 5. Then the root is found within 5 iterations as follows -
X[1] = 4.34636336183652
X[2] = 4.13470839381104
X[3] = 4.1110189242975
X[4] = 4.1107172220963
X[5] = 4.11071702042904
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Let the second estimate be x = -5. Then the root is found within 6 iterations
X[1] = -3.71323014398166
X[2] = -3.16489091626675
X[3] = -3.05219606148819
X[4] = -3.04755119034318
X[5] = -3.0475450433433
X[6] = -3.0475450455597
_______________________________________…
Let the final estimate be x = 0. The root is found within 5 iterations
X[1] = 1.63599196873366
X[2] = 1.4289257999748
X[3] = 1.43681987521638
X[4] = 1.43682802645487
X[5] = 1.4368280252333
All 3 roots have been found. They are x = 4.1107 , x= -3.0475, x= 1.4368.
In newton's method you have to make initial estimate as a guess. Since we knew 3 roots exist, we took one positive value, one negative value and one value as zero.
We keep computing newton's method with various estimates until all the roots are found.
You can learn about Newton's method at
http://en.wikipedia.org/wiki/Newton%27s_…
However, this is not the only method. Their exist more methods such as bisection method, false position, etc etc
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METHOD 2.
we know 3 roots exist since the degree is 3. Let those 3 roots be a, b, c.
We divide both sides of the equation by 2 and get
x^3 - 2.5x^2 - 11x + 18 = 0
i.e.
(x+a)(x+b)(x+c) = x^3 - 2.5x^2 - 11x + 18
x^3+x^2*c+x^2*b+x*b*c+a*x^2+a*x*c+a*b*x… = x^3 - 2.5x^2 - 11x + 18
x^2(c+b+a) + x(bc+ac+ab) + abc = - 2.5x^2 - 11x + 18
now we see that,
c+b+a = -2.5
bc+ac+ab = -11
abc = 18
using linear algebra and other manipulations you should end up with a = 4.1107 , b= -3.0475, c= 1.4368. These are the same values computed in method 1.
Since degree here is 3, this method can be applied. But if its a large degree than this method is not practical. This is also an alternate to quadratic formula and works easily
There will be 3 solutions since the degree is 3.
Let initial x = 5. Then the root is found within 5 iterations as follows -
X[1] = 4.34636336183652
X[2] = 4.13470839381104
X[3] = 4.1110189242975
X[4] = 4.1107172220963
X[5] = 4.11071702042904
__________________________________
Let the second estimate be x = -5. Then the root is found within 6 iterations
X[1] = -3.71323014398166
X[2] = -3.16489091626675
X[3] = -3.05219606148819
X[4] = -3.04755119034318
X[5] = -3.0475450433433
X[6] = -3.0475450455597
_______________________________________…
Let the final estimate be x = 0. The root is found within 5 iterations
X[1] = 1.63599196873366
X[2] = 1.4289257999748
X[3] = 1.43681987521638
X[4] = 1.43682802645487
X[5] = 1.4368280252333
All 3 roots have been found. They are x = 4.1107 , x= -3.0475, x= 1.4368.
In newton's method you have to make initial estimate as a guess. Since we knew 3 roots exist, we took one positive value, one negative value and one value as zero.
We keep computing newton's method with various estimates until all the roots are found.
You can learn about Newton's method at
http://en.wikipedia.org/wiki/Newton%27s_…
However, this is not the only method. Their exist more methods such as bisection method, false position, etc etc
_______________________________________…
METHOD 2.
we know 3 roots exist since the degree is 3. Let those 3 roots be a, b, c.
We divide both sides of the equation by 2 and get
x^3 - 2.5x^2 - 11x + 18 = 0
i.e.
(x+a)(x+b)(x+c) = x^3 - 2.5x^2 - 11x + 18
x^3+x^2*c+x^2*b+x*b*c+a*x^2+a*x*c+a*b*x… = x^3 - 2.5x^2 - 11x + 18
x^2(c+b+a) + x(bc+ac+ab) + abc = - 2.5x^2 - 11x + 18
now we see that,
c+b+a = -2.5
bc+ac+ab = -11
abc = 18
using linear algebra and other manipulations you should end up with a = 4.1107 , b= -3.0475, c= 1.4368. These are the same values computed in method 1.
Since degree here is 3, this method can be applied. But if its a large degree than this method is not practical. This is also an alternate to quadratic formula and works easily
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PEMDAS