Solve lambda(lambda-4)(lambda-3)+3lambda-9=0

$$\lambda ( \lambda - 4 ) ( \lambda - 3 ) + 3 \lambda - 9 = 0$$

Solve for λ

$$\lambda =3$$

$$\lambda =1$$

Solution Steps

Multiply and combine like terms.

$$\lambda ^{3}-7\lambda ^{2}+15\lambda -9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}-7\lambda ^{2}+15\lambda -9 by \lambda -1 to get \lambda ^{2}-6\lambda +9. Solve the equation where the result equals to 0.

$$\lambda ^{2}-6\lambda +9=0$$

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, -6 for b, and 9 for c in the quadratic formula.

$$\lambda =\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$\lambda =\frac{6±0}{2}$$

Solutions are the same.

$$\lambda =3$$

List all found solutions.

$$\lambda =1 \lambda =3$$