Solve -16x^2-24x-5=

$$- 16 x ^ { 2 } - 24 x - 5 =$$

Factor

$$-\left(4x+1\right)\left(4x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -16x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-24 ab=-16\left(-5\right)=80$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -24.

$$a=-4 b=-20$$

Rewrite -16x^{2}-24x-5 as \left(-16x^{2}-4x\right)+\left(-20x-5\right).

$$\left(-16x^{2}-4x\right)+\left(-20x-5\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(-4x-1\right)+5\left(-4x-1\right)$$

Factor out common term -4x-1 by using distributive property.

$$\left(-4x-1\right)\left(4x+5\right)$$

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.

$$-16x^{2}-24x-5=0$$

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.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-16\right)\left(-5\right)}}{2\left(-16\right)}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\left(-16\right)\left(-5\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-\left(-24\right)±\sqrt{576+64\left(-5\right)}}{2\left(-16\right)}$$

Multiply 64 times -5.

$$x=\frac{-\left(-24\right)±\sqrt{576-320}}{2\left(-16\right)}$$

Add 576 to -320.

$$x=\frac{-\left(-24\right)±\sqrt{256}}{2\left(-16\right)}$$

Take the square root of 256.

$$x=\frac{-\left(-24\right)±16}{2\left(-16\right)}$$

The opposite of -24 is 24.

$$x=\frac{24±16}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{24±16}{-32}$$

Now solve the equation x=\frac{24±16}{-32} when ± is plus. Add 24 to 16.

$$x=\frac{40}{-32}$$

Reduce the fraction \frac{40}{-32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{5}{4}$$

Now solve the equation x=\frac{24±16}{-32} when ± is minus. Subtract 16 from 24.

$$x=\frac{8}{-32}$$

Reduce the fraction \frac{8}{-32} to lowest terms by extracting and canceling out 8.

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

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

$$-16x^{2}-24x-5=-16\left(x-\left(-\frac{5}{4}\right)\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

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

$$-16x^{2}-24x-5=-16\left(x+\frac{5}{4}\right)\left(x+\frac{1}{4}\right)$$

Add \frac{5}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}-24x-5=-16\times \frac{-4x-5}{-4}\left(x+\frac{1}{4}\right)$$

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

$$-16x^{2}-24x-5=-16\times \frac{-4x-5}{-4}\times \frac{-4x-1}{-4}$$

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

$$-16x^{2}-24x-5=-16\times \frac{\left(-4x-5\right)\left(-4x-1\right)}{-4\left(-4\right)}$$

Multiply -4 times -4.

$$-16x^{2}-24x-5=-16\times \frac{\left(-4x-5\right)\left(-4x-1\right)}{16}$$

Cancel out 16, the greatest common factor in -16 and 16.

$$-16x^{2}-24x-5=-\left(-4x-5\right)\left(-4x-1\right)$$

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.

$$x ^ 2 +\frac{3}{2}x +\frac{5}{16} = 0$$

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

$$r + s = -\frac{3}{2} rs = \frac{5}{16}$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}. 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>

$$r = -\frac{3}{4} - u s = -\frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{16}

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = \frac{5}{16}$$

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

$$\frac{9}{16} - u^2 = \frac{5}{16}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = \frac{5}{16}-\frac{9}{16} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =-\frac{3}{4} - \frac{1}{2} = -1.250 s = -\frac{3}{4} + \frac{1}{2} = -0.250$$

Evaluate

$$-16x^{2}-24x-5$$