Solve 3x^2+18x-48

$$3 x ^ { 2 } + 18 x - 48$$

Factor

$$3\left(x-2\right)\left(x+8\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}+6x-16\right)$$

Consider x^{2}+6x-16. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-16\right)=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite x^{2}+6x-16 as \left(x^{2}-2x\right)+\left(8x-16\right).

$$\left(x^{2}-2x\right)+\left(8x-16\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-2\right)+8\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+8\right)$$

Rewrite the complete factored expression.

$$3\left(x-2\right)\left(x+8\right)$$

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.

$$3x^{2}+18x-48=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\left(-48\right)}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\left(-48\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\left(-48\right)}}{2\times 3}$$

Multiply -12 times -48.

$$x=\frac{-18±\sqrt{324+576}}{2\times 3}$$

Add 324 to 576.

$$x=\frac{-18±\sqrt{900}}{2\times 3}$$

Take the square root of 900.

$$x=\frac{-18±30}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±30}{6}$$

Now solve the equation x=\frac{-18±30}{6} when ± is plus. Add -18 to 30.

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

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{-18±30}{6} when ± is minus. Subtract 30 from -18.

$$x=-\frac{48}{6}$$

Divide -48 by 6.

$$x=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -8 for x_{2}.

$$3x^{2}+18x-48=3\left(x-2\right)\left(x-\left(-8\right)\right)$$

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

$$3x^{2}+18x-48=3\left(x-2\right)\left(x+8\right)$$

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

$$x ^ 2 +6x -16 = 0$$

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

$$r + s = -6 rs = -16$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -16

$$(-3 - u) (-3 + u) = -16$$

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

$$9 - u^2 = -16$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -16-9 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =-3 - 5 = -8 s = -3 + 5 = 2$$

Evaluate

$$3\left(x-2\right)\left(x+8\right)$$