Find the roots of the polynomial equation.
x3−2x2+10x+136=0
(1 point)
Responses
−3±i, 4
−3±i, 4
3±i, 4
3±i, 4
3±5i, −4
3±5i, −4
−3±5i, −4
To find the roots of the polynomial equation, we can use various methods like factoring, synthetic division, or using the quadratic formula. In this case, none of those methods are immediately apparent, so we can resort to using a numerical method called the Newton-Raphson method.
Using the Newton-Raphson method, we start with an initial guess for a root and then iterate through an equation to get closer and closer to the actual root.
Start by choosing an initial guess, let's say x = 0. We can calculate the equation and its derivative at this point:
f(x) = x^3 - 2x^2 + 10x + 136
f'(x) = 3x^2 - 4x + 10
Using these values, we can apply the Newton-Raphson iteration formula:
x1 = x0 - f(x0)/f'(x0)
where x1 is a better approximation for the root and x0 is the initial guess. Repeat this process until we find a satisfactory approximation for the root.
Using x0 = 0, we get:
x1 = 0 - [0^3 - 2(0)^2 + 10(0) + 136] / [3(0)^2 - 4(0) + 10]
= -136 / 10
= -13.6
Now, we can repeat the iteration process with x0 = -13.6:
x1 = -13.6 - [-13.6^3 - 2(-13.6)^2 + 10(-13.6) + 136] / [3(-13.6)^2 - 4(-13.6) + 10]
= -13.6 - (-21499.1) / (986.56 + 54.4 + 10)
= -13.6 + 21.81 / 1051.96
= -13.6 + 0.021 / 1051.96
= -13.6 + 0.00002
≈ -13.6
We repeat this process until we get a satisfactory approximation for the root. In this case, after iterating a few more times, we find that the root is approximately -13.6.
Therefore, one of the roots of the polynomial equation x^3 - 2x^2 + 10x + 136 = 0 is x ≈ -13.6.
Now, we can use synthetic division (dividing by x + 13.6) to factoring out the polynomial equation:
x^3 - 2x^2 + 10x + 136 = (x + 13.6)(x^2 - 15.6x + 9.44)
Now, we can solve the quadratic equation x^2 - 15.6x + 9.44 = 0 using the quadratic formula:
x = (-(-15.6) ± √((-15.6)^2 - 4(1)(9.44))) / (2(1))
≈ (15.6 ± √(243.36 - 37.76)) / 2
≈ (15.6 ± √205.6) / 2
≈ (15.6 ± 14.34) / 2
Therefore, the other roots of the polynomial equation are approximately:
x ≈ (15.6 + 14.34) / 2 ≈ 29.94 / 2 ≈ 14.97
x ≈ (15.6 - 14.34) / 2 ≈ 1.26 / 2 ≈ 0.63
So, the roots of the polynomial equation x^3 - 2x^2 + 10x + 136 = 0 are approximately -13.6, 14.97, and 0.63.