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

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

Factor

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

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.

$$a+b=2 ab=1\left(-3\right)=-3$$

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.

$$a=-1 b=3$$

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

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

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

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

Factor out common term x-1 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square 10.

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

Multiply -4 times 5.

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

Multiply -20 times -15.

$$x=\frac{-10±\sqrt{100+300}}{2\times 5}$$

Add 100 to 300.

$$x=\frac{-10±\sqrt{400}}{2\times 5}$$

Take the square root of 400.

$$x=\frac{-10±20}{2\times 5}$$

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}.

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

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.This is achieved by dividing both sides of the equation by 5

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

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.

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

Evaluate

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