Solve 4k^2-8k-5

$$4 k ^ { 2 } - 8 k - 5$$

Factor

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

Solution Steps

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

$$a+b=-8 ab=4\left(-5\right)=-20$$

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 4k^{2}-8k-5 as \left(4k^{2}-10k\right)+\left(2k-5\right).

$$\left(4k^{2}-10k\right)+\left(2k-5\right)$$

Factor out 2k in 4k^{2}-10k.

$$2k\left(2k-5\right)+2k-5$$

Factor out common term 2k-5 by using distributive property.

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

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.

$$4k^{2}-8k-5=0$$

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.

$$k=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square -8.

$$k=\frac{-\left(-8\right)±\sqrt{64-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$k=\frac{-\left(-8\right)±\sqrt{64-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$k=\frac{-\left(-8\right)±\sqrt{64+80}}{2\times 4}$$

Add 64 to 80.

$$k=\frac{-\left(-8\right)±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$k=\frac{-\left(-8\right)±12}{2\times 4}$$

The opposite of -8 is 8.

$$k=\frac{8±12}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{8±12}{8}$$

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

$$k=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$k=\frac{5}{2}$$

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

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

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$k=-\frac{1}{2}$$

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

$$4k^{2}-8k-5=4\left(k-\frac{5}{2}\right)\left(k-\left(-\frac{1}{2}\right)\right)$$

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

$$4k^{2}-8k-5=4\left(k-\frac{5}{2}\right)\left(k+\frac{1}{2}\right)$$

Subtract \frac{5}{2} from k by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4k^{2}-8k-5=4\times \frac{2k-5}{2}\left(k+\frac{1}{2}\right)$$

Add \frac{1}{2} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4k^{2}-8k-5=4\times \frac{2k-5}{2}\times \frac{2k+1}{2}$$

Multiply \frac{2k-5}{2} times \frac{2k+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4k^{2}-8k-5=4\times \frac{\left(2k-5\right)\left(2k+1\right)}{2\times 2}$$

Multiply 2 times 2.

$$4k^{2}-8k-5=4\times \frac{\left(2k-5\right)\left(2k+1\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4k^{2}-8k-5=\left(2k-5\right)\left(2k+1\right)$$

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 4

$$x ^ 2 -2x -\frac{5}{4} = 0$$

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

$$r + s = 2 rs = -\frac{5}{4}$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{4}

$$(1 - u) (1 + u) = -\frac{5}{4}$$

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

$$1 - u^2 = -\frac{5}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{5}{4}-1 = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =1 - \frac{3}{2} = -0.500 s = 1 + \frac{3}{2} = 2.500$$

Evaluate

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