Solve 3z^2-14z-24

$$3 z ^ { 2 } - 14 z - 24$$

Factor

$$\left(z-6\right)\left(3z+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3z^{2}+az+bz-24. To find a and b, set up a system to be solved.

$$a+b=-14 ab=3\left(-24\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -14.

$$a=-18 b=4$$

Rewrite 3z^{2}-14z-24 as \left(3z^{2}-18z\right)+\left(4z-24\right).

$$\left(3z^{2}-18z\right)+\left(4z-24\right)$$

Factor out 3z in the first and 4 in the second group.

$$3z\left(z-6\right)+4\left(z-6\right)$$

Factor out common term z-6 by using distributive property.

$$\left(z-6\right)\left(3z+4\right)$$

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.

$$3z^{2}-14z-24=0$$

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.

$$z=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

Square -14.

$$z=\frac{-\left(-14\right)±\sqrt{196-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$z=\frac{-\left(-14\right)±\sqrt{196-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$z=\frac{-\left(-14\right)±\sqrt{196+288}}{2\times 3}$$

Add 196 to 288.

$$z=\frac{-\left(-14\right)±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$z=\frac{-\left(-14\right)±22}{2\times 3}$$

The opposite of -14 is 14.

$$z=\frac{14±22}{2\times 3}$$

Multiply 2 times 3.

$$z=\frac{14±22}{6}$$

Now solve the equation z=\frac{14±22}{6} when ± is plus. Add 14 to 22.

$$z=\frac{36}{6}$$

Divide 36 by 6.

$$z=6$$

Now solve the equation z=\frac{14±22}{6} when ± is minus. Subtract 22 from 14.

$$z=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$z=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{4}{3} for x_{2}.

$$3z^{2}-14z-24=3\left(z-6\right)\left(z-\left(-\frac{4}{3}\right)\right)$$

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

$$3z^{2}-14z-24=3\left(z-6\right)\left(z+\frac{4}{3}\right)$$

Add \frac{4}{3} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3z^{2}-14z-24=3\left(z-6\right)\times \frac{3z+4}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3z^{2}-14z-24=\left(z-6\right)\left(3z+4\right)$$

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

$$x ^ 2 -\frac{14}{3}x -8 = 0$$

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

$$r + s = \frac{14}{3} rs = -8$$

Two numbers r and s sum up to \frac{14}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{14}{3} = \frac{7}{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>

$$r = \frac{7}{3} - u s = \frac{7}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -8

$$(\frac{7}{3} - u) (\frac{7}{3} + u) = -8$$

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

$$\frac{49}{9} - u^2 = -8$$

Simplify the expression by subtracting \frac{49}{9} on both sides

$$-u^2 = -8-\frac{49}{9} = -\frac{121}{9}$$

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

$$u^2 = \frac{121}{9} u = \pm\sqrt{\frac{121}{9}} = \pm \frac{11}{3}$$

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

$$r =\frac{7}{3} - \frac{11}{3} = -1.333 s = \frac{7}{3} + \frac{11}{3} = 6$$

Evaluate

$$\left(z-6\right)\left(3z+4\right)$$