Solve -(lambda-1)(lambda^2-blambda+5)=0

$$- ( \lambda - 1 ) ( \lambda ^ { 2 } - b \lambda + 5 ) = 0$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\lambda +\frac{5}{\lambda }\text{, }&\lambda \neq 0\\b\in \mathrm{C}\text{, }&\lambda =1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of \lambda -1, find the opposite of each term.

$$\left(-\lambda +1\right)\left(\lambda ^{2}-b\lambda +5\right)=0$$

Use the distributive property to multiply -\lambda +1 by \lambda ^{2}-b\lambda +5.

$$-\lambda ^{3}+\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=0$$

Add \lambda ^{3} to both sides. Anything plus zero gives itself.

$$\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=\lambda ^{3}$$

Add 5\lambda to both sides.

$$\lambda ^{2}b+\lambda ^{2}-b\lambda +5=\lambda ^{3}+5\lambda$$

Subtract \lambda ^{2} from both sides.

$$\lambda ^{2}b-b\lambda +5=\lambda ^{3}+5\lambda -\lambda ^{2}$$

Subtract 5 from both sides.

$$\lambda ^{2}b-b\lambda =\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

Combine all terms containing b.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

The equation is in standard form.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}-\lambda ^{2}+5\lambda -5$$

Divide both sides by \lambda ^{2}-\lambda .

$$\frac{\left(\lambda ^{2}-\lambda \right)b}{\lambda ^{2}-\lambda }=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

Dividing by \lambda ^{2}-\lambda undoes the multiplication by \lambda ^{2}-\lambda .

$$b=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

Divide \left(-1+\lambda \right)\left(5+\lambda ^{2}\right) by \lambda ^{2}-\lambda .

$$b=\lambda +\frac{5}{\lambda }$$

Solve for b

$$\left\{\begin{matrix}b=\lambda +\frac{5}{\lambda }\text{, }&\lambda \neq 0\\b\in \mathrm{R}\text{, }&\lambda =1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of \lambda -1, find the opposite of each term.

$$\left(-\lambda +1\right)\left(\lambda ^{2}-b\lambda +5\right)=0$$

Use the distributive property to multiply -\lambda +1 by \lambda ^{2}-b\lambda +5.

$$-\lambda ^{3}+\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=0$$

Add \lambda ^{3} to both sides. Anything plus zero gives itself.

$$\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=\lambda ^{3}$$

Add 5\lambda to both sides.

$$\lambda ^{2}b+\lambda ^{2}-b\lambda +5=\lambda ^{3}+5\lambda$$

Subtract \lambda ^{2} from both sides.

$$\lambda ^{2}b-b\lambda +5=\lambda ^{3}+5\lambda -\lambda ^{2}$$

Subtract 5 from both sides.

$$\lambda ^{2}b-b\lambda =\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

Combine all terms containing b.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

The equation is in standard form.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}-\lambda ^{2}+5\lambda -5$$

Divide both sides by \lambda ^{2}-\lambda .

$$\frac{\left(\lambda ^{2}-\lambda \right)b}{\lambda ^{2}-\lambda }=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

Dividing by \lambda ^{2}-\lambda undoes the multiplication by \lambda ^{2}-\lambda .

$$b=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

Divide \left(-1+\lambda \right)\left(5+\lambda ^{2}\right) by \lambda ^{2}-\lambda .

$$b=\lambda +\frac{5}{\lambda }$$

Solve for λ (complex solution)

$$\lambda =\frac{-\sqrt{b^{2}-20}+b}{2}$$

$$\lambda =1$$

$$\lambda =\frac{\sqrt{b^{2}-20}+b}{2}$$

Solve for λ

$$\left\{\begin{matrix}\\\lambda =1\text{, }&\text{unconditionally}\\\lambda =\frac{\sqrt{b^{2}-20}+b}{2}\text{; }\lambda =\frac{-\sqrt{b^{2}-20}+b}{2}\text{, }&|b|\geq 2\sqrt{5}\end{matrix}\right.$$