Solve d^2-4d-5

$$d ^ { 2 } - 4 d - 5$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as d^{2}+ad+bd-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 d^{2}-4d-5 as \left(d^{2}-5d\right)+\left(d-5\right).

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

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

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

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

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

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.

$$d^{2}-4d-5=0$$

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.

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

Square -4.

$$d=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$d=\frac{-\left(-4\right)±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$d=\frac{-\left(-4\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$d=\frac{-\left(-4\right)±6}{2}$$

The opposite of -4 is 4.

$$d=\frac{4±6}{2}$$

Now solve the equation d=\frac{4±6}{2} when ± is plus. Add 4 to 6.

$$d=\frac{10}{2}$$

Divide 10 by 2.

$$d=5$$

Now solve the equation d=\frac{4±6}{2} when ± is minus. Subtract 6 from 4.

$$d=-\frac{2}{2}$$

Divide -2 by 2.

$$d=-1$$

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}.

$$d^{2}-4d-5=\left(d-5\right)\left(d-\left(-1\right)\right)$$

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

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

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.

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

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

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