Solve r^2-3r+2

$$r ^ { 2 } - 3 r + 2$$

Factor

$$\left(r-2\right)\left(r-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br+2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

Factor out r in the first and -1 in the second group.

$$r\left(r-2\right)-\left(r-2\right)$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(r-1\right)$$

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.

$$r^{2}-3r+2=0$$

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.

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

Square -3.

$$r=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$r=\frac{-\left(-3\right)±\sqrt{9-8}}{2}$$

Add 9 to -8.

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

Take the square root of 1.

$$r=\frac{-\left(-3\right)±1}{2}$$

The opposite of -3 is 3.

$$r=\frac{3±1}{2}$$

Now solve the equation r=\frac{3±1}{2} when ± is plus. Add 3 to 1.

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

Divide 4 by 2.

$$r=2$$

Now solve the equation r=\frac{3±1}{2} when ± is minus. Subtract 1 from 3.

$$r=\frac{2}{2}$$

Divide 2 by 2.

$$r=1$$

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

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -3x +2 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 3 rs = 2$$

Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{3}{2} - u s = \frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(\frac{3}{2} - u) (\frac{3}{2} + u) = 2$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{4} - u^2 = 2$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = 2-\frac{9}{4} = -\frac{1}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{3}{2} - \frac{1}{2} = 1 s = \frac{3}{2} + \frac{1}{2} = 2$$

Evaluate

$$\left(r-2\right)\left(r-1\right)$$