Solve f(x)=-x^2+2x+15

$$f ( x ) = - x ^ { 2 } + 2 x + 15$$

Factor

$$-\left(x-5\right)\left(x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=2 ab=-15=-15$$

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. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=5 b=-3$$

Rewrite -x^{2}+2x+15 as \left(-x^{2}+5x\right)+\left(-3x+15\right).

$$\left(-x^{2}+5x\right)+\left(-3x+15\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-3\right)$$

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.

$$-x^{2}+2x+15=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 15}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 15}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 15}}{2\left(-1\right)}$$

Multiply 4 times 15.

$$x=\frac{-2±\sqrt{4+60}}{2\left(-1\right)}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-2±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±8}{-2}$$

Now solve the equation x=\frac{-2±8}{-2} when ± is plus. Add -2 to 8.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{-2±8}{-2} when ± is minus. Subtract 8 from -2.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

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 5 for x_{2}.

$$-x^{2}+2x+15=-\left(x-\left(-3\right)\right)\left(x-5\right)$$

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

$$-x^{2}+2x+15=-\left(x+3\right)\left(x-5\right)$$

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.

$$x ^ 2 -2x -15 = 0$$

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

$$r + s = 2 rs = -15$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -15

$$(1 - u) (1 + u) = -15$$

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

$$1 - u^2 = -15$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -15-1 = -16$$

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

$$u^2 = 16 u = \pm\sqrt{16} = \pm 4$$

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

$$r =1 - 4 = -3 s = 1 + 4 = 5$$

Evaluate

$$-\left(x-5\right)\left(x+3\right)$$