Solve 6a^2-18a-24

$$6 a ^ { 2 } - 18 a - 24$$

Factor

$$6\left(a-4\right)\left(a+1\right)$$

Solution Steps

Factor out 6.

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

Consider a^{2}-3a-4. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-4. To find p and q, set up a system to be solved.

$$p+q=-3 pq=1\left(-4\right)=-4$$

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 -4.

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

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$p=-4 q=1$$

Rewrite a^{2}-3a-4 as \left(a^{2}-4a\right)+\left(a-4\right).

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

Factor out a in a^{2}-4a.

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

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

$$\left(a-4\right)\left(a+1\right)$$

Rewrite the complete factored expression.

$$6\left(a-4\right)\left(a+1\right)$$

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.

$$6a^{2}-18a-24=0$$

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.

$$a=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 6\left(-24\right)}}{2\times 6}$$

Square -18.

$$a=\frac{-\left(-18\right)±\sqrt{324-4\times 6\left(-24\right)}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-\left(-18\right)±\sqrt{324-24\left(-24\right)}}{2\times 6}$$

Multiply -24 times -24.

$$a=\frac{-\left(-18\right)±\sqrt{324+576}}{2\times 6}$$

Add 324 to 576.

$$a=\frac{-\left(-18\right)±\sqrt{900}}{2\times 6}$$

Take the square root of 900.

$$a=\frac{-\left(-18\right)±30}{2\times 6}$$

The opposite of -18 is 18.

$$a=\frac{18±30}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{18±30}{12}$$

Now solve the equation a=\frac{18±30}{12} when ± is plus. Add 18 to 30.

$$a=\frac{48}{12}$$

Divide 48 by 12.

$$a=4$$

Now solve the equation a=\frac{18±30}{12} when ± is minus. Subtract 30 from 18.

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

Divide -12 by 12.

$$a=-1$$

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 -1 for x_{2}.

$$6a^{2}-18a-24=6\left(a-4\right)\left(a-\left(-1\right)\right)$$

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

$$6a^{2}-18a-24=6\left(a-4\right)\left(a+1\right)$$

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 6

$$x ^ 2 -3x -4 = 0$$

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

$$r + s = 3 rs = -4$$

Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(\frac{3}{2} - u) (\frac{3}{2} + u) = -4$$

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

$$\frac{9}{4} - u^2 = -4$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -4-\frac{9}{4} = -\frac{25}{4}$$

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.

$$r =\frac{3}{2} - \frac{5}{2} = -1 s = \frac{3}{2} + \frac{5}{2} = 4$$

Evaluate

$$6\left(a-4\right)\left(a+1\right)$$