Solve -2x^2-10x+12

$$- 2 x ^ { 2 } - 10 x + 12$$

Factor

$$2\left(1-x\right)\left(x+6\right)$$

Solution Steps

Factor out 2.

$$2\left(-x^{2}-5x+6\right)$$

Consider -x^{2}-5x+6. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -x^{2}-5x+6 as \left(-x^{2}+x\right)+\left(-6x+6\right).

$$\left(-x^{2}+x\right)+\left(-6x+6\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(-x+1\right)+6\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$2\left(-x+1\right)\left(x+6\right)$$

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.

$$-2x^{2}-10x+12=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\times 12}}{2\left(-2\right)}$$

Multiply 8 times 12.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\left(-2\right)}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\left(-2\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±14}{-4}$$

Now solve the equation x=\frac{10±14}{-4} when ± is plus. Add 10 to 14.

$$x=\frac{24}{-4}$$

Divide 24 by -4.

$$x=-6$$

Now solve the equation x=\frac{10±14}{-4} when ± is minus. Subtract 14 from 10.

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

Divide -4 by -4.

$$x=1$$

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 1 for x_{2}.

$$-2x^{2}-10x+12=-2\left(x-\left(-6\right)\right)\left(x-1\right)$$

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

$$-2x^{2}-10x+12=-2\left(x+6\right)\left(x-1\right)$$

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.

$$x ^ 2 +5x -6 = 0$$

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

$$r + s = -5 rs = -6$$

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{2}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{5}{2} - u s = -\frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -6

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -6$$

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

$$\frac{25}{4} - u^2 = -6$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -6-\frac{25}{4} = -\frac{49}{4}$$

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

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

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

$$r =-\frac{5}{2} - \frac{7}{2} = -6 s = -\frac{5}{2} + \frac{7}{2} = 1$$

Evaluate

$$2\left(1-x\right)\left(x+6\right)$$