Solve 2x^2-6x-80=

$$2 x ^ { 2 } - 6 x - 80 =$$

Factor

$$2\left(x-8\right)\left(x+5\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{2}-3x-40\right)$$

Consider x^{2}-3x-40. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-40\right)=-40$$

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 -40.

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

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite x^{2}-3x-40 as \left(x^{2}-8x\right)+\left(5x-40\right).

$$\left(x^{2}-8x\right)+\left(5x-40\right)$$

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

$$x\left(x-8\right)+5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$2\left(x-8\right)\left(x+5\right)$$

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.

$$2x^{2}-6x-80=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-80\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-80\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-80\right)}}{2\times 2}$$

Multiply -8 times -80.

$$x=\frac{-\left(-6\right)±\sqrt{36+640}}{2\times 2}$$

Add 36 to 640.

$$x=\frac{-\left(-6\right)±\sqrt{676}}{2\times 2}$$

Take the square root of 676.

$$x=\frac{-\left(-6\right)±26}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±26}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±26}{4}$$

Now solve the equation x=\frac{6±26}{4} when ± is plus. Add 6 to 26.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{6±26}{4} when ± is minus. Subtract 26 from 6.

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

Divide -20 by 4.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -5 for x_{2}.

$$2x^{2}-6x-80=2\left(x-8\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}-6x-80=2\left(x-8\right)\left(x+5\right)$$

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

$$x ^ 2 -3x -40 = 0$$

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

$$r + s = 3 rs = -40$$

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

$$r = \frac{3}{2} - u s = \frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -40

$$(\frac{3}{2} - u) (\frac{3}{2} + u) = -40$$

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

$$\frac{9}{4} - u^2 = -40$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -40-\frac{9}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =\frac{3}{2} - \frac{13}{2} = -5 s = \frac{3}{2} + \frac{13}{2} = 8$$

Evaluate

$$2\left(x-8\right)\left(x+5\right)$$