Solve 3x^2+60x-288

$$3 x ^ { 2 } + 60 x - 288$$

Factor

$$3\left(x-4\right)\left(x+24\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}+20x-96\right)$$

Consider x^{2}+20x-96. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\left(-96\right)=-96$$

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 -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 20.

$$a=-4 b=24$$

Rewrite x^{2}+20x-96 as \left(x^{2}-4x\right)+\left(24x-96\right).

$$\left(x^{2}-4x\right)+\left(24x-96\right)$$

Factor out x in the first and 24 in the second group.

$$x\left(x-4\right)+24\left(x-4\right)$$

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

$$\left(x-4\right)\left(x+24\right)$$

Rewrite the complete factored expression.

$$3\left(x-4\right)\left(x+24\right)$$

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.

$$3x^{2}+60x-288=0$$

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.

$$x=\frac{-60±\sqrt{60^{2}-4\times 3\left(-288\right)}}{2\times 3}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 3\left(-288\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-60±\sqrt{3600-12\left(-288\right)}}{2\times 3}$$

Multiply -12 times -288.

$$x=\frac{-60±\sqrt{3600+3456}}{2\times 3}$$

Add 3600 to 3456.

$$x=\frac{-60±\sqrt{7056}}{2\times 3}$$

Take the square root of 7056.

$$x=\frac{-60±84}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-60±84}{6}$$

Now solve the equation x=\frac{-60±84}{6} when ± is plus. Add -60 to 84.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{-60±84}{6} when ± is minus. Subtract 84 from -60.

$$x=-\frac{144}{6}$$

Divide -144 by 6.

$$x=-24$$

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 -24 for x_{2}.

$$3x^{2}+60x-288=3\left(x-4\right)\left(x-\left(-24\right)\right)$$

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

$$3x^{2}+60x-288=3\left(x-4\right)\left(x+24\right)$$

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

$$x ^ 2 +20x -96 = 0$$

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

$$r + s = -20 rs = -96$$

Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -10. 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>

$$r = -10 - u s = -10 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -96

$$(-10 - u) (-10 + u) = -96$$

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

$$100 - u^2 = -96$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = -96-100 = -196$$

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

$$u^2 = 196 u = \pm\sqrt{196} = \pm 14$$

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

$$r =-10 - 14 = -24 s = -10 + 14 = 4$$

Evaluate

$$3\left(x-4\right)\left(x+24\right)$$