Solve 2k^2-16k-40

$$2 k ^ { 2 } - 16 k - 40$$

Factor

$$2\left(k-10\right)\left(k+2\right)$$

Solution Steps

Factor out 2.

$$2\left(k^{2}-8k-20\right)$$

Consider k^{2}-8k-20. Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk-20. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-20\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite k^{2}-8k-20 as \left(k^{2}-10k\right)+\left(2k-20\right).

$$\left(k^{2}-10k\right)+\left(2k-20\right)$$

Factor out k in the first and 2 in the second group.

$$k\left(k-10\right)+2\left(k-10\right)$$

Factor out common term k-10 by using distributive property.

$$\left(k-10\right)\left(k+2\right)$$

Rewrite the complete factored expression.

$$2\left(k-10\right)\left(k+2\right)$$

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.

$$2k^{2}-16k-40=0$$

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.

$$k=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

Square -16.

$$k=\frac{-\left(-16\right)±\sqrt{256-4\times 2\left(-40\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-16\right)±\sqrt{256-8\left(-40\right)}}{2\times 2}$$

Multiply -8 times -40.

$$k=\frac{-\left(-16\right)±\sqrt{256+320}}{2\times 2}$$

Add 256 to 320.

$$k=\frac{-\left(-16\right)±\sqrt{576}}{2\times 2}$$

Take the square root of 576.

$$k=\frac{-\left(-16\right)±24}{2\times 2}$$

The opposite of -16 is 16.

$$k=\frac{16±24}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{16±24}{4}$$

Now solve the equation k=\frac{16±24}{4} when ± is plus. Add 16 to 24.

$$k=\frac{40}{4}$$

Divide 40 by 4.

$$k=10$$

Now solve the equation k=\frac{16±24}{4} when ± is minus. Subtract 24 from 16.

$$k=-\frac{8}{4}$$

Divide -8 by 4.

$$k=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -2 for x_{2}.

$$2k^{2}-16k-40=2\left(k-10\right)\left(k-\left(-2\right)\right)$$

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

$$2k^{2}-16k-40=2\left(k-10\right)\left(k+2\right)$$

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

$$x ^ 2 -8x -20 = 0$$

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

$$r + s = 8 rs = -20$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -20

$$(4 - u) (4 + u) = -20$$

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

$$16 - u^2 = -20$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -20-16 = -36$$

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

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

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

$$r =4 - 6 = -2 s = 4 + 6 = 10$$

Evaluate

$$2\left(k-10\right)\left(k+2\right)$$