Solve 5a^2+5a-30

$$5 a ^ { 2 } + 5 a - 30$$

Factor

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

Solution Steps

Factor out 5.

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

Consider a^{2}+a-6. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-6. To find p and q, set up a system to be solved.

$$p+q=1 pq=1\left(-6\right)=-6$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. 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.

$$p=-2 q=3$$

Rewrite a^{2}+a-6 as \left(a^{2}-2a\right)+\left(3a-6\right).

$$\left(a^{2}-2a\right)+\left(3a-6\right)$$

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

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

Factor out common term a-2 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$5a^{2}+5a-30=0$$

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.

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

Square 5.

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

Multiply -4 times 5.

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

Multiply -20 times -30.

$$a=\frac{-5±\sqrt{25+600}}{2\times 5}$$

Add 25 to 600.

$$a=\frac{-5±\sqrt{625}}{2\times 5}$$

Take the square root of 625.

$$a=\frac{-5±25}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{-5±25}{10}$$

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

$$a=\frac{20}{10}$$

Divide 20 by 10.

$$a=2$$

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

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

Divide -30 by 10.

$$a=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -3 for x_{2}.

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

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

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

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

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

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.

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

Evaluate

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