Solve =3-2lambda-lambda^2

$$= 3 - 2 \lambda - \lambda ^ { 2 }$$

Factor

$$\left(1-\lambda \right)\left(\lambda +3\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-\lambda ^{2}-2\lambda +3$$

Factor the expression by grouping. First, the expression needs to be rewritten as -\lambda ^{2}+a\lambda +b\lambda +3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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.

$$a=1 b=-3$$

Rewrite -\lambda ^{2}-2\lambda +3 as \left(-\lambda ^{2}+\lambda \right)+\left(-3\lambda +3\right).

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

Factor out \lambda in the first and 3 in the second group.

$$\lambda \left(-\lambda +1\right)+3\left(-\lambda +1\right)$$

Factor out common term -\lambda +1 by using distributive property.

$$\left(-\lambda +1\right)\left(\lambda +3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square -2.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-1\right)}$$

Add 4 to 12.

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

Take the square root of 16.

$$\lambda =\frac{-\left(-2\right)±4}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$\lambda =\frac{2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$\lambda =\frac{2±4}{-2}$$

Now solve the equation \lambda =\frac{2±4}{-2} when ± is plus. Add 2 to 4.

$$\lambda =\frac{6}{-2}$$

Divide 6 by -2.

$$\lambda =-3$$

Now solve the equation \lambda =\frac{2±4}{-2} when ± is minus. Subtract 4 from 2.

$$\lambda =-\frac{2}{-2}$$

Divide -2 by -2.

$$\lambda =1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and 1 for x_{2}.

$$-\lambda ^{2}-2\lambda +3=-\left(\lambda -\left(-3\right)\right)\left(\lambda -1\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-\lambda ^{2}-2\lambda +3=-\left(\lambda +3\right)\left(\lambda -1\right)$$

Evaluate

$$\left(1-\lambda \right)\left(\lambda +3\right)$$