Solve -6x^2-14x-8

$$- 6 x ^ { 2 } - 14 x - 8$$

Factor

$$2\left(-x-1\right)\left(3x+4\right)$$

Solution Steps

Factor out 2.

$$2\left(-3x^{2}-7x-4\right)$$

Consider -3x^{2}-7x-4. Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-3\left(-4\right)=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-3 b=-4$$

Rewrite -3x^{2}-7x-4 as \left(-3x^{2}-3x\right)+\left(-4x-4\right).

$$\left(-3x^{2}-3x\right)+\left(-4x-4\right)$$

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

$$3x\left(-x-1\right)+4\left(-x-1\right)$$

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

$$\left(-x-1\right)\left(3x+4\right)$$

Rewrite the complete factored expression.

$$2\left(-x-1\right)\left(3x+4\right)$$

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.

$$-6x^{2}-14x-8=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-14\right)±\sqrt{196+24\left(-8\right)}}{2\left(-6\right)}$$

Multiply 24 times -8.

$$x=\frac{-\left(-14\right)±\sqrt{196-192}}{2\left(-6\right)}$$

Add 196 to -192.

$$x=\frac{-\left(-14\right)±\sqrt{4}}{2\left(-6\right)}$$

Take the square root of 4.

$$x=\frac{-\left(-14\right)±2}{2\left(-6\right)}$$

The opposite of -14 is 14.

$$x=\frac{14±2}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{14±2}{-12}$$

Now solve the equation x=\frac{14±2}{-12} when ± is plus. Add 14 to 2.

$$x=\frac{16}{-12}$$

Reduce the fraction \frac{16}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{14±2}{-12} when ± is minus. Subtract 2 from 14.

$$x=\frac{12}{-12}$$

Divide 12 by -12.

$$x=-1$$

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

$$-6x^{2}-14x-8=-6\left(x-\left(-\frac{4}{3}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$-6x^{2}-14x-8=-6\left(x+\frac{4}{3}\right)\left(x+1\right)$$

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

$$-6x^{2}-14x-8=-6\times \frac{-3x-4}{-3}\left(x+1\right)$$

Cancel out 3, the greatest common factor in -6 and 3.

$$-6x^{2}-14x-8=2\left(-3x-4\right)\left(x+1\right)$$

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.

$$x ^ 2 +\frac{7}{3}x +\frac{4}{3} = 0$$

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

$$r + s = -\frac{7}{3} rs = \frac{4}{3}$$

Two numbers r and s sum up to -\frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{3} = -\frac{7}{6}. 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>

$$r = -\frac{7}{6} - u s = -\frac{7}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{3}

$$(-\frac{7}{6} - u) (-\frac{7}{6} + u) = \frac{4}{3}$$

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

$$\frac{49}{36} - u^2 = \frac{4}{3}$$

Simplify the expression by subtracting \frac{49}{36} on both sides

$$-u^2 = \frac{4}{3}-\frac{49}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =-\frac{7}{6} - \frac{1}{6} = -1.333 s = -\frac{7}{6} + \frac{1}{6} = -1.000$$

Evaluate

$$-6x^{2}-14x-8$$