Solve -3x^2-16x-5

$$- 3 x ^ { 2 } - 16 x - 5$$

Factor

$$-\left(x+5\right)\left(3x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=-3\left(-5\right)=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-1 b=-15$$

Rewrite -3x^{2}-16x-5 as \left(-3x^{2}-x\right)+\left(-15x-5\right).

$$\left(-3x^{2}-x\right)+\left(-15x-5\right)$$

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

$$-x\left(3x+1\right)-5\left(3x+1\right)$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(-x-5\right)$$

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.

$$-3x^{2}-16x-5=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-16\right)±\sqrt{256+12\left(-5\right)}}{2\left(-3\right)}$$

Multiply 12 times -5.

$$x=\frac{-\left(-16\right)±\sqrt{256-60}}{2\left(-3\right)}$$

Add 256 to -60.

$$x=\frac{-\left(-16\right)±\sqrt{196}}{2\left(-3\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-16\right)±14}{2\left(-3\right)}$$

The opposite of -16 is 16.

$$x=\frac{16±14}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{16±14}{-6}$$

Now solve the equation x=\frac{16±14}{-6} when ± is plus. Add 16 to 14.

$$x=\frac{30}{-6}$$

Divide 30 by -6.

$$x=-5$$

Now solve the equation x=\frac{16±14}{-6} when ± is minus. Subtract 14 from 16.

$$x=\frac{2}{-6}$$

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

$$x=-\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and -\frac{1}{3} for x_{2}.

$$-3x^{2}-16x-5=-3\left(x-\left(-5\right)\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$-3x^{2}-16x-5=-3\left(x+5\right)\left(x+\frac{1}{3}\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}-16x-5=-3\left(x+5\right)\times \frac{-3x-1}{-3}$$

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

$$-3x^{2}-16x-5=\left(x+5\right)\left(-3x-1\right)$$

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.

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

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

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

Two numbers r and s sum up to -\frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{3} = -\frac{8}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(-\frac{8}{3} - u) (-\frac{8}{3} + u) = \frac{5}{3}$$

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

$$\frac{64}{9} - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting \frac{64}{9} on both sides

$$-u^2 = \frac{5}{3}-\frac{64}{9} = -\frac{49}{9}$$

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

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

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

$$r =-\frac{8}{3} - \frac{7}{3} = -5 s = -\frac{8}{3} + \frac{7}{3} = -0.333$$

Evaluate

$$-\left(x+5\right)\left(3x+1\right)$$