Solve lambda^3-12lambda^2+36lambda-32=0

$$\lambda ^ { 3 } - 12 \lambda ^ { 2 } + 36 \lambda - 32 = 0$$

Solve for λ

$$\lambda =2$$

$$\lambda =8$$

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 -32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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.

$$\lambda =2$$

By Factor theorem, \lambda -k is a factor of the polynomial for each root k. Divide \lambda ^{3}-12\lambda ^{2}+36\lambda -32 by \lambda -2 to get \lambda ^{2}-10\lambda +16. Solve the equation where the result equals to 0.

$$\lambda ^{2}-10\lambda +16=0$$

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, -10 for b, and 16 for c in the quadratic formula.

$$\lambda =\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$\lambda =\frac{10±6}{2}$$

Solve the equation \lambda ^{2}-10\lambda +16=0 when ± is plus and when ± is minus.

$$\lambda =2 \lambda =8$$

List all found solutions.

$$\lambda =2 \lambda =8$$