Solve h(x)=-5x^2+10x+15

$$h ( x ) = - 5 x ^ { 2 } + 10 x + 15$$

Factor

$$-5\left(x-3\right)\left(x+1\right)$$

Solution Steps

Factor out 5.

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

Consider -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.

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

Rewrite the complete factored expression.

$$5\left(x-3\right)\left(-x-1\right)$$

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.

$$-5x^{2}+10x+15=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-5\right)\times 15}}{2\left(-5\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-5\right)\times 15}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-10±\sqrt{100+20\times 15}}{2\left(-5\right)}$$

Multiply 20 times 15.

$$x=\frac{-10±\sqrt{100+300}}{2\left(-5\right)}$$

Add 100 to 300.

$$x=\frac{-10±\sqrt{400}}{2\left(-5\right)}$$

Take the square root of 400.

$$x=\frac{-10±20}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-10±20}{-10}$$

Now solve the equation x=\frac{-10±20}{-10} when ± is plus. Add -10 to 20.

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

Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{-10±20}{-10} when ± is minus. Subtract 20 from -10.

$$x=-\frac{30}{-10}$$

Divide -30 by -10.

$$x=3$$

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}.

$$-5x^{2}+10x+15=-5\left(x-\left(-1\right)\right)\left(x-3\right)$$

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

$$-5x^{2}+10x+15=-5\left(x+1\right)\left(x-3\right)$$

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.

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

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

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

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 = -3

$$(1 - u) (1 + u) = -3$$

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

$$1 - u^2 = -3$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -3-1 = -4$$

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

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

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

$$r =1 - 2 = -1 s = 1 + 2 = 3$$

Evaluate

$$-5\left(x-3\right)\left(x+1\right)$$