Solve lambda^3+4lambda^2+5lambda+2=0Rightarrow

$$\lambda ^ { 3 } + 4 \lambda ^ { 2 } + 5 \lambda + 2 = 0 \Rightarrow$$

Solve for λ

$$\lambda =-1$$

$$\lambda =-2$$

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 2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$\lambda =-1$$

By Factor theorem, \lambda -k is a factor of the polynomial for each root k. Divide \lambda ^{3}+4\lambda ^{2}+5\lambda +2 by \lambda +1 to get \lambda ^{2}+3\lambda +2. Solve the equation where the result equals to 0.

$$\lambda ^{2}+3\lambda +2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 3 for b, and 2 for c in the quadratic formula.

$$\lambda =\frac{-3±\sqrt{3^{2}-4\times 1\times 2}}{2}$$

Do the calculations.

$$\lambda =\frac{-3±1}{2}$$

Solve the equation \lambda ^{2}+3\lambda +2=0 when ± is plus and when ± is minus.

$$\lambda =-2 \lambda =-1$$

List all found solutions.

$$\lambda =-1 \lambda =-2$$