Solve F(x)=-x^2+5x-6

$$F ( x ) = - x ^ { 2 } + 5 x - 6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 -x^{2}+5x-6 as \left(-x^{2}+3x\right)+\left(2x-6\right).

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

Factor out -x in the first and 2 in the second group.

$$-x\left(x-3\right)+2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+2\right)$$

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.

$$-x^{2}+5x-6=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-5±\sqrt{25-24}}{2\left(-1\right)}$$

Add 25 to -24.

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

Take the square root of 1.

$$x=\frac{-5±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±1}{-2}$$

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

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

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

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

Divide -6 by -2.

$$x=3$$

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 3 for x_{2}.

$$-x^{2}+5x-6=-\left(x-2\right)\left(x-3\right)$$

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.

$$x ^ 2 -5x +6 = 0$$

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

$$r + s = 5 rs = 6$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 6

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = 6$$

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

$$\frac{25}{4} - u^2 = 6$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = 6-\frac{25}{4} = -\frac{1}{4}$$

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.

$$r =\frac{5}{2} - \frac{1}{2} = 2 s = \frac{5}{2} + \frac{1}{2} = 3$$

Evaluate

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