Solve 3a^2-3a-60

$$3 a ^ { 2 } - 3 a - 60$$

Factor

$$3\left(a-5\right)\left(a+4\right)$$

Solution Steps

Factor out 3.

$$3\left(a^{2}-a-20\right)$$

Consider a^{2}-a-20. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-20. To find p and q, set up a system to be solved.

$$p+q=-1 pq=1\left(-20\right)=-20$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$p=-5 q=4$$

Rewrite a^{2}-a-20 as \left(a^{2}-5a\right)+\left(4a-20\right).

$$\left(a^{2}-5a\right)+\left(4a-20\right)$$

Factor out a in the first and 4 in the second group.

$$a\left(a-5\right)+4\left(a-5\right)$$

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

$$\left(a-5\right)\left(a+4\right)$$

Rewrite the complete factored expression.

$$3\left(a-5\right)\left(a+4\right)$$

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.

$$3a^{2}-3a-60=0$$

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.

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-60\right)}}{2\times 3}$$

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-60\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-3\right)±\sqrt{9-12\left(-60\right)}}{2\times 3}$$

Multiply -12 times -60.

$$a=\frac{-\left(-3\right)±\sqrt{9+720}}{2\times 3}$$

Add 9 to 720.

$$a=\frac{-\left(-3\right)±\sqrt{729}}{2\times 3}$$

Take the square root of 729.

$$a=\frac{-\left(-3\right)±27}{2\times 3}$$

The opposite of -3 is 3.

$$a=\frac{3±27}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{3±27}{6}$$

Now solve the equation a=\frac{3±27}{6} when ± is plus. Add 3 to 27.

$$a=\frac{30}{6}$$

Divide 30 by 6.

$$a=5$$

Now solve the equation a=\frac{3±27}{6} when ± is minus. Subtract 27 from 3.

$$a=-\frac{24}{6}$$

Divide -24 by 6.

$$a=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -4 for x_{2}.

$$3a^{2}-3a-60=3\left(a-5\right)\left(a-\left(-4\right)\right)$$

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

$$3a^{2}-3a-60=3\left(a-5\right)\left(a+4\right)$$

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 3

$$x ^ 2 -1x -20 = 0$$

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

$$r + s = 1 rs = -20$$

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 = -20

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -20$$

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

$$\frac{1}{4} - u^2 = -20$$

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

$$-u^2 = -20-\frac{1}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =\frac{1}{2} - \frac{9}{2} = -4 s = \frac{1}{2} + \frac{9}{2} = 5$$

Evaluate

$$3\left(a-5\right)\left(a+4\right)$$