Solve -3x^2-48x+108

$$- 3 x ^ { 2 } - 48 x + 108$$

Factor

$$3\left(2-x\right)\left(x+18\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}-16x+36\right)$$

Consider -x^{2}-16x+36. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-16 ab=-36=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -16.

$$a=2 b=-18$$

Rewrite -x^{2}-16x+36 as \left(-x^{2}+2x\right)+\left(-18x+36\right).

$$\left(-x^{2}+2x\right)+\left(-18x+36\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(-x+2\right)+18\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(x+18\right)$$

Rewrite the complete factored expression.

$$3\left(-x+2\right)\left(x+18\right)$$

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.

$$-3x^{2}-48x+108=0$$

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.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\left(-3\right)\times 108}}{2\left(-3\right)}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\left(-3\right)\times 108}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-48\right)±\sqrt{2304+12\times 108}}{2\left(-3\right)}$$

Multiply 12 times 108.

$$x=\frac{-\left(-48\right)±\sqrt{2304+1296}}{2\left(-3\right)}$$

Add 2304 to 1296.

$$x=\frac{-\left(-48\right)±\sqrt{3600}}{2\left(-3\right)}$$

Take the square root of 3600.

$$x=\frac{-\left(-48\right)±60}{2\left(-3\right)}$$

The opposite of -48 is 48.

$$x=\frac{48±60}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{48±60}{-6}$$

Now solve the equation x=\frac{48±60}{-6} when ± is plus. Add 48 to 60.

$$x=\frac{108}{-6}$$

Divide 108 by -6.

$$x=-18$$

Now solve the equation x=\frac{48±60}{-6} when ± is minus. Subtract 60 from 48.

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

Divide -12 by -6.

$$x=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -18 for x_{1} and 2 for x_{2}.

$$-3x^{2}-48x+108=-3\left(x-\left(-18\right)\right)\left(x-2\right)$$

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

$$-3x^{2}-48x+108=-3\left(x+18\right)\left(x-2\right)$$

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.

$$x ^ 2 +16x -36 = 0$$

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

$$r + s = -16 rs = -36$$

Two numbers r and s sum up to -16 exactly when the average of the two numbers is \frac{1}{2}*-16 = -8. 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>

$$r = -8 - u s = -8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -36

$$(-8 - u) (-8 + u) = -36$$

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

$$64 - u^2 = -36$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = -36-64 = -100$$

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

$$u^2 = 100 u = \pm\sqrt{100} = \pm 10$$

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

$$r =-8 - 10 = -18 s = -8 + 10 = 2$$

Evaluate

$$3\left(2-x\right)\left(x+18\right)$$