solve for x: factor (-x^3)+(2x^2)+(x)-26

Question

linear algebra

Answer 1

Test for rational zeroes first. Factors of the constant 26 are ±1, ±2, ±13, ±26.

Setting x to these in turn does not give us a zero expression, but, because the polynomial is a cubic, there must be at least one real zero.

If f(x)=-x³+2x²+x-26, then f(-3)=16 and f(-2)=-12, so there is a zero when -3<x<-2. Newton's Method provides an iterative way to find the zero, using x_n+1=x_n-f(x_n)/f'(x_n).

f'(x)=-3x²+4x+1. Using the intermediate Value Theorem we can start the process by x₀=(-3-2)/2=-2.5.

x₁=-2.5135..., x₂=-2.51345..., ..., x=-2.5134514815737 approx.

Represent this zero by X, and use synthetic division to find the quadratic factor:

X | -3    2               1     -26

     -3 -3X       2X-3X² |  26

     -3 2-3X 1+2X-3X² |  0 = -3x²+(2-3X)x+(1+2X-3X²).

Let a=-3, b=2-3X, c=1+2X-3X², then:

x=(-b±√(b²-4ac))/2a.

b=9.540354 approx, c=-22.979218 approx.

4ac=275.750616 approx, b²=91.018363 approx, so b²-4ac<0.

x=(-b±i√(4ac-b²))/2a=(-9.540354±i√184.732253)/-6.

x=1.5901±2.2653i approx are the complex zeroes.