Solve k^2+5k+4

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

Factor

$$\left(k+1\right)\left(k+4\right)$$

Solution Steps

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

$$a+b=5 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

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

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

Factor out k in the first and 4 in the second group.

$$k\left(k+1\right)+4\left(k+1\right)$$

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

$$\left(k+1\right)\left(k+4\right)$$

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.

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

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.

$$k=\frac{-5±\sqrt{5^{2}-4\times 4}}{2}$$

Square 5.

$$k=\frac{-5±\sqrt{25-4\times 4}}{2}$$

Multiply -4 times 4.

$$k=\frac{-5±\sqrt{25-16}}{2}$$

Add 25 to -16.

$$k=\frac{-5±\sqrt{9}}{2}$$

Take the square root of 9.

$$k=\frac{-5±3}{2}$$

Now solve the equation k=\frac{-5±3}{2} when ± is plus. Add -5 to 3.

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

Divide -2 by 2.

$$k=-1$$

Now solve the equation k=\frac{-5±3}{2} when ± is minus. Subtract 3 from -5.

$$k=-\frac{8}{2}$$

Divide -8 by 2.

$$k=-4$$

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 -4 for x_{2}.

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

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

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

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.

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

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

$$r + s = -5 rs = 4$$

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = 4

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = 4$$

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

$$\frac{25}{4} - u^2 = 4$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = 4-\frac{25}{4} = -\frac{9}{4}$$

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.

$$r =-\frac{5}{2} - \frac{3}{2} = -4 s = -\frac{5}{2} + \frac{3}{2} = -1$$

Evaluate

$$\left(k+1\right)\left(k+4\right)$$