Solve 12x^2+18x-54

$$12 x ^ { 2 } + 18 x - 54$$

Factor

$$6\left(2x-3\right)\left(x+3\right)$$

Solution Steps

Factor out 6.

$$6\left(2x^{2}+3x-9\right)$$

Consider 2x^{2}+3x-9. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-9\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite 2x^{2}+3x-9 as \left(2x^{2}-3x\right)+\left(6x-9\right).

$$\left(2x^{2}-3x\right)+\left(6x-9\right)$$

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

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

Factor out common term 2x-3 by using distributive property.

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

Rewrite the complete factored expression.

$$6\left(2x-3\right)\left(x+3\right)$$

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.

$$12x^{2}+18x-54=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 12\left(-54\right)}}{2\times 12}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 12\left(-54\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-18±\sqrt{324-48\left(-54\right)}}{2\times 12}$$

Multiply -48 times -54.

$$x=\frac{-18±\sqrt{324+2592}}{2\times 12}$$

Add 324 to 2592.

$$x=\frac{-18±\sqrt{2916}}{2\times 12}$$

Take the square root of 2916.

$$x=\frac{-18±54}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-18±54}{24}$$

Now solve the equation x=\frac{-18±54}{24} when ± is plus. Add -18 to 54.

$$x=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{-18±54}{24} when ± is minus. Subtract 54 from -18.

$$x=-\frac{72}{24}$$

Divide -72 by 24.

$$x=-3$$

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

$$12x^{2}+18x-54=12\left(x-\frac{3}{2}\right)\left(x-\left(-3\right)\right)$$

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

$$12x^{2}+18x-54=12\left(x-\frac{3}{2}\right)\left(x+3\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+18x-54=12\times \frac{2x-3}{2}\left(x+3\right)$$

Cancel out 2, the greatest common factor in 12 and 2.

$$12x^{2}+18x-54=6\left(2x-3\right)\left(x+3\right)$$

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 12

$$x ^ 2 +\frac{3}{2}x -\frac{9}{2} = 0$$

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

$$r + s = -\frac{3}{2} rs = -\frac{9}{2}$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}. 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>

$$r = -\frac{3}{4} - u s = -\frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{2}

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -\frac{9}{2}$$

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

$$\frac{9}{16} - u^2 = -\frac{9}{2}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{9}{2}-\frac{9}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{3}{4} - \frac{9}{4} = -3 s = -\frac{3}{4} + \frac{9}{4} = 1.500$$

Evaluate

$$6\left(2x-3\right)\left(x+3\right)$$