Solve f(x)=3+2x-x^2

$$f ( x ) = 3 + 2 x - x ^ { 2 }$$

Factor

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

Solution Steps

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

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

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

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

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

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

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

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

$$\left(x-3\right)\left(-x-1\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 3.

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

Add 4 to 12.

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

Take the square root of 16.

$$x=\frac{-2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±4}{-2}$$

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

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

Divide 2 by -2.

$$x=-1$$

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

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

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 -1 for x_{1} and 3 for x_{2}.

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

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

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

Evaluate

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