Solve 5x^2-5x-30

$$5 x ^ { 2 } - 5 x - 30$$

Factor

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

Solution Steps

Factor out 5.

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

Consider x^{2}-x-6. 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.

$$a+b=-1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite x^{2}-x-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.

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

Rewrite the complete factored expression.

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

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.

$$5x^{2}-5x-30=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-30\right)}}{2\times 5}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 5\left(-30\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-20\left(-30\right)}}{2\times 5}$$

Multiply -20 times -30.

$$x=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 5}$$

Add 25 to 600.

$$x=\frac{-\left(-5\right)±\sqrt{625}}{2\times 5}$$

Take the square root of 625.

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

The opposite of -5 is 5.

$$x=\frac{5±25}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{5±25}{10}$$

Now solve the equation x=\frac{5±25}{10} when ± is plus. Add 5 to 25.

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

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{5±25}{10} when ± is minus. Subtract 25 from 5.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

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 -2 for x_{2}.

$$5x^{2}-5x-30=5\left(x-3\right)\left(x-\left(-2\right)\right)$$

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

$$5x^{2}-5x-30=5\left(x-3\right)\left(x+2\right)$$

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

$$x ^ 2 -1x -6 = 0$$

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

$$r + s = 1 rs = -6$$

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

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

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

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -6$$

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

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

Simplify the expression by subtracting \frac{1}{4} on both sides

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

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

$$u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{2}$$

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

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

Evaluate

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