Solve 24z^2-14z-5

$$24 z ^ { 2 } - 14 z - 5$$

Factor

$$\left(6z-5\right)\left(4z+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24z^{2}+az+bz-5. To find a and b, set up a system to be solved.

$$a+b=-14 ab=24\left(-5\right)=-120$$

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 -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -14.

$$a=-20 b=6$$

Rewrite 24z^{2}-14z-5 as \left(24z^{2}-20z\right)+\left(6z-5\right).

$$\left(24z^{2}-20z\right)+\left(6z-5\right)$$

Factor out 4z in 24z^{2}-20z.

$$4z\left(6z-5\right)+6z-5$$

Factor out common term 6z-5 by using distributive property.

$$\left(6z-5\right)\left(4z+1\right)$$

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.

$$24z^{2}-14z-5=0$$

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.

$$z=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 24\left(-5\right)}}{2\times 24}$$

Square -14.

$$z=\frac{-\left(-14\right)±\sqrt{196-4\times 24\left(-5\right)}}{2\times 24}$$

Multiply -4 times 24.

$$z=\frac{-\left(-14\right)±\sqrt{196-96\left(-5\right)}}{2\times 24}$$

Multiply -96 times -5.

$$z=\frac{-\left(-14\right)±\sqrt{196+480}}{2\times 24}$$

Add 196 to 480.

$$z=\frac{-\left(-14\right)±\sqrt{676}}{2\times 24}$$

Take the square root of 676.

$$z=\frac{-\left(-14\right)±26}{2\times 24}$$

The opposite of -14 is 14.

$$z=\frac{14±26}{2\times 24}$$

Multiply 2 times 24.

$$z=\frac{14±26}{48}$$

Now solve the equation z=\frac{14±26}{48} when ± is plus. Add 14 to 26.

$$z=\frac{40}{48}$$

Reduce the fraction \frac{40}{48} to lowest terms by extracting and canceling out 8.

$$z=\frac{5}{6}$$

Now solve the equation z=\frac{14±26}{48} when ± is minus. Subtract 26 from 14.

$$z=-\frac{12}{48}$$

Reduce the fraction \frac{-12}{48} to lowest terms by extracting and canceling out 12.

$$z=-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and -\frac{1}{4} for x_{2}.

$$24z^{2}-14z-5=24\left(z-\frac{5}{6}\right)\left(z-\left(-\frac{1}{4}\right)\right)$$

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

$$24z^{2}-14z-5=24\left(z-\frac{5}{6}\right)\left(z+\frac{1}{4}\right)$$

Subtract \frac{5}{6} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24z^{2}-14z-5=24\times \frac{6z-5}{6}\left(z+\frac{1}{4}\right)$$

Add \frac{1}{4} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24z^{2}-14z-5=24\times \frac{6z-5}{6}\times \frac{4z+1}{4}$$

Multiply \frac{6z-5}{6} times \frac{4z+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24z^{2}-14z-5=24\times \frac{\left(6z-5\right)\left(4z+1\right)}{6\times 4}$$

Multiply 6 times 4.

$$24z^{2}-14z-5=24\times \frac{\left(6z-5\right)\left(4z+1\right)}{24}$$

Cancel out 24, the greatest common factor in 24 and 24.

$$24z^{2}-14z-5=\left(6z-5\right)\left(4z+1\right)$$

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 24

$$x ^ 2 -\frac{7}{12}x -\frac{5}{24} = 0$$

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

$$r + s = \frac{7}{12} rs = -\frac{5}{24}$$

Two numbers r and s sum up to \frac{7}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{12} = \frac{7}{24}. 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>

$$r = \frac{7}{24} - u s = \frac{7}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{24}

$$(\frac{7}{24} - u) (\frac{7}{24} + u) = -\frac{5}{24}$$

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

$$\frac{49}{576} - u^2 = -\frac{5}{24}$$

Simplify the expression by subtracting \frac{49}{576} on both sides

$$-u^2 = -\frac{5}{24}-\frac{49}{576} = -\frac{169}{576}$$

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

$$u^2 = \frac{169}{576} u = \pm\sqrt{\frac{169}{576}} = \pm \frac{13}{24}$$

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

$$r =\frac{7}{24} - \frac{13}{24} = -0.250 s = \frac{7}{24} + \frac{13}{24} = 0.833$$

Evaluate

$$\left(6z-5\right)\left(4z+1\right)$$