Solve (x_0^3-3x_0^2+4)

$$( x _ { 0 } ^ { 3 } - 3 x _ { 0 } ^ { 2 } + 4 )$$

Factor

$$\left(x_{0}+1\right)\left(x_{0}-2\right)^{2}$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. One such root is 2. Factor the polynomial by dividing it by x_{0}-2.

$$\left(x_{0}-2\right)\left(x_{0}^{2}-x_{0}-2\right)$$

Consider x_{0}^{2}-x_{0}-2. Factor the expression by grouping. First, the expression needs to be rewritten as x_{0}^{2}+ax_{0}+bx_{0}-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite x_{0}^{2}-x_{0}-2 as \left(x_{0}^{2}-2x_{0}\right)+\left(x_{0}-2\right).

$$\left(x_{0}^{2}-2x_{0}\right)+\left(x_{0}-2\right)$$

Factor out x_{0} in x_{0}^{2}-2x_{0}.

$$x_{0}\left(x_{0}-2\right)+x_{0}-2$$

Factor out common term x_{0}-2 by using distributive property.

$$\left(x_{0}-2\right)\left(x_{0}+1\right)$$

Rewrite the complete factored expression.

$$\left(x_{0}+1\right)\left(x_{0}-2\right)^{2}$$

Evaluate

$$\left(x_{0}+1\right)\left(x_{0}-2\right)^{2}$$