Solve 12x^2-24x-63

$$12 x ^ { 2 } - 24 x - 63$$

Factor

$$3\left(2x-7\right)\left(2x+3\right)$$

Solution Steps

Factor out 3.

$$3\left(4x^{2}-8x-21\right)$$

Consider 4x^{2}-8x-21. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=-8 ab=4\left(-21\right)=-84$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -84.

$$1,-84 2,-42 3,-28 4,-21 6,-14 7,-12$$

Calculate the sum for each pair.

$$1-84=-83 2-42=-40 3-28=-25 4-21=-17 6-14=-8 7-12=-5$$

The solution is the pair that gives sum -8.

$$a=-14 b=6$$

Rewrite 4x^{2}-8x-21 as \left(4x^{2}-14x\right)+\left(6x-21\right).

$$\left(4x^{2}-14x\right)+\left(6x-21\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(2x-7\right)+3\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(2x+3\right)$$

Rewrite the complete factored expression.

$$3\left(2x-7\right)\left(2x+3\right)$$

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.

$$12x^{2}-24x-63=0$$

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.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 12\left(-63\right)}}{2\times 12}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 12\left(-63\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-24\right)±\sqrt{576-48\left(-63\right)}}{2\times 12}$$

Multiply -48 times -63.

$$x=\frac{-\left(-24\right)±\sqrt{576+3024}}{2\times 12}$$

Add 576 to 3024.

$$x=\frac{-\left(-24\right)±\sqrt{3600}}{2\times 12}$$

Take the square root of 3600.

$$x=\frac{-\left(-24\right)±60}{2\times 12}$$

The opposite of -24 is 24.

$$x=\frac{24±60}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{24±60}{24}$$

Now solve the equation x=\frac{24±60}{24} when ± is plus. Add 24 to 60.

$$x=\frac{84}{24}$$

Reduce the fraction \frac{84}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{24±60}{24} when ± is minus. Subtract 60 from 24.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and -\frac{3}{2} for x_{2}.

$$12x^{2}-24x-63=12\left(x-\frac{7}{2}\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$12x^{2}-24x-63=12\left(x-\frac{7}{2}\right)\left(x+\frac{3}{2}\right)$$

Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-24x-63=12\times \frac{2x-7}{2}\left(x+\frac{3}{2}\right)$$

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

$$12x^{2}-24x-63=12\times \frac{2x-7}{2}\times \frac{2x+3}{2}$$

Multiply \frac{2x-7}{2} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-24x-63=12\times \frac{\left(2x-7\right)\left(2x+3\right)}{2\times 2}$$

Multiply 2 times 2.

$$12x^{2}-24x-63=12\times \frac{\left(2x-7\right)\left(2x+3\right)}{4}$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12x^{2}-24x-63=3\left(2x-7\right)\left(2x+3\right)$$

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

$$x ^ 2 -2x -\frac{21}{4} = 0$$

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

$$r + s = 2 rs = -\frac{21}{4}$$

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 = -\frac{21}{4}

$$(1 - u) (1 + u) = -\frac{21}{4}$$

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

$$1 - u^2 = -\frac{21}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{21}{4}-1 = -\frac{25}{4}$$

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.

$$r =1 - \frac{5}{2} = -1.500 s = 1 + \frac{5}{2} = 3.500$$

Evaluate

$$12x^{2}-24x-63$$