Solve 3a^2-6a-24

$$3 a ^ { 2 } - 6 a - 24$$

Factor

$$3\left(a-4\right)\left(a+2\right)$$

Solution Steps

Factor out 3.

$$3\left(a^{2}-2a-8\right)$$

Consider a^{2}-2a-8. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-8. To find p and q, set up a system to be solved.

$$p+q=-2 pq=1\left(-8\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$p=-4 q=2$$

Rewrite a^{2}-2a-8 as \left(a^{2}-4a\right)+\left(2a-8\right).

$$\left(a^{2}-4a\right)+\left(2a-8\right)$$

Factor out a in the first and 2 in the second group.

$$a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(a+2\right)$$

Rewrite the complete factored expression.

$$3\left(a-4\right)\left(a+2\right)$$

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.

$$3a^{2}-6a-24=0$$

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.

$$a=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

Square -6.

$$a=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-6\right)±\sqrt{36-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$a=\frac{-\left(-6\right)±\sqrt{36+288}}{2\times 3}$$

Add 36 to 288.

$$a=\frac{-\left(-6\right)±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$a=\frac{-\left(-6\right)±18}{2\times 3}$$

The opposite of -6 is 6.

$$a=\frac{6±18}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{6±18}{6}$$

Now solve the equation a=\frac{6±18}{6} when ± is plus. Add 6 to 18.

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

Divide 24 by 6.

$$a=4$$

Now solve the equation a=\frac{6±18}{6} when ± is minus. Subtract 18 from 6.

$$a=-\frac{12}{6}$$

Divide -12 by 6.

$$a=-2$$

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

$$3a^{2}-6a-24=3\left(a-4\right)\left(a-\left(-2\right)\right)$$

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

$$3a^{2}-6a-24=3\left(a-4\right)\left(a+2\right)$$

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 3

$$x ^ 2 -2x -8 = 0$$

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

$$r + s = 2 rs = -8$$

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 = -8

$$(1 - u) (1 + u) = -8$$

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

$$1 - u^2 = -8$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -8-1 = -9$$

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

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

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

$$r =1 - 3 = -2 s = 1 + 3 = 4$$

Evaluate

$$3\left(a-4\right)\left(a+2\right)$$