Solve k^2+4k-5

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

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk-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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

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

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

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

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

Factor out common term k-1 by using distributive property.

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

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.

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

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.

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

Square 4.

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

Multiply -4 times -5.

$$k=\frac{-4±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$k=\frac{-4±\sqrt{36}}{2}$$

Take the square root of 36.

$$k=\frac{-4±6}{2}$$

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

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

Divide 2 by 2.

$$k=1$$

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

$$k=-\frac{10}{2}$$

Divide -10 by 2.

$$k=-5$$

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

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

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

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

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.

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

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.

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

Evaluate

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