Solve (4-lambda)(1-lambda)+2=0

$$( 4 - \lambda ) ( 1 - \lambda ) + 2 = 0$$

Solve for λ

$$\lambda =3$$

$$\lambda =2$$

Steps Using Factoring

Use the distributive property to multiply 4-\lambda by 1-\lambda and combine like terms.

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

Add 4 and 2 to get 6.

$$6-5\lambda +\lambda ^{2}=0$$

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

$$\lambda ^{2}-5\lambda +6=0$$

To solve the equation, factor \lambda ^{2}-5\lambda +6 using formula \lambda ^{2}+\left(a+b\right)\lambda +ab=\left(\lambda +a\right)\left(\lambda +b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite factored expression \left(\lambda +a\right)\left(\lambda +b\right) using the obtained values.

$$\left(\lambda -3\right)\left(\lambda -2\right)$$

To find equation solutions, solve \lambda -3=0 and \lambda -2=0.

$$\lambda =3 \lambda =2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 4-\lambda by 1-\lambda and combine like terms.

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

Add 4 and 2 to get 6.

$$6-5\lambda +\lambda ^{2}=0$$

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

$$\lambda ^{2}-5\lambda +6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as \lambda ^{2}+a\lambda +b\lambda +6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite \lambda ^{2}-5\lambda +6 as \left(\lambda ^{2}-3\lambda \right)+\left(-2\lambda +6\right).

$$\left(\lambda ^{2}-3\lambda \right)+\left(-2\lambda +6\right)$$

Factor out \lambda in the first and -2 in the second group.

$$\lambda \left(\lambda -3\right)-2\left(\lambda -3\right)$$

Factor out common term \lambda -3 by using distributive property.

$$\left(\lambda -3\right)\left(\lambda -2\right)$$

To find equation solutions, solve \lambda -3=0 and \lambda -2=0.

$$\lambda =3 \lambda =2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4-\lambda by 1-\lambda and combine like terms.

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

Add 4 and 2 to get 6.

$$6-5\lambda +\lambda ^{2}=0$$

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

$$\lambda ^{2}-5\lambda +6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$\lambda =\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$\lambda =\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$\lambda =\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$\lambda =\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$\lambda =\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$\lambda =\frac{5±1}{2}$$

Now solve the equation \lambda =\frac{5±1}{2} when ± is plus. Add 5 to 1.

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

Divide 6 by 2.

$$\lambda =3$$

Now solve the equation \lambda =\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

$$\lambda =\frac{4}{2}$$

Divide 4 by 2.

$$\lambda =2$$

The equation is now solved.

$$\lambda =3 \lambda =2$$

Steps for Completing the Square

Use the distributive property to multiply 4-\lambda by 1-\lambda and combine like terms.

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

Add 4 and 2 to get 6.

$$6-5\lambda +\lambda ^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-5\lambda +\lambda ^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\lambda ^{2}-5\lambda =-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\lambda ^{2}-5\lambda +\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$\lambda ^{2}-5\lambda +\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$\lambda ^{2}-5\lambda +\frac{25}{4}=\frac{1}{4}$$

Factor \lambda ^{2}-5\lambda +\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(\lambda -\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\lambda -\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$\lambda -\frac{5}{2}=\frac{1}{2} \lambda -\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$\lambda =3 \lambda =2$$