Solve =2x^2-8x-10

$$= 2 x ^ { 2 } - 8 x - 10$$

Factor

$$2\left(x-5\right)\left(x+1\right)$$

Solution Steps

Factor out 2.

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

Consider x^{2}-4x-5. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

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

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

Factor out x in x^{2}-5x.

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

Factor out common term x-5 by using distributive property.

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

Rewrite the complete factored expression.

$$2\left(x-5\right)\left(x+1\right)$$

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.

$$2x^{2}-8x-10=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\times 2}$$

Add 64 to 80.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

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

The opposite of -8 is 8.

$$x=\frac{8±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±12}{4}$$

Now solve the equation x=\frac{8±12}{4} when ± is plus. Add 8 to 12.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{8±12}{4} when ± is minus. Subtract 12 from 8.

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

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 5 for x_{1} and -1 for x_{2}.

$$2x^{2}-8x-10=2\left(x-5\right)\left(x-\left(-1\right)\right)$$

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

$$2x^{2}-8x-10=2\left(x-5\right)\left(x+1\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 -4x -5 = 0$$

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

$$r + s = 4 rs = -5$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(2 - u) (2 + u) = -5$$

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

$$4 - u^2 = -5$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -5-4 = -9$$

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

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

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

$$r =2 - 3 = -1 s = 2 + 3 = 5$$

Evaluate

$$2\left(x-5\right)\left(x+1\right)$$