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a+b=5 ab=1\times 4=4
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.
1,4 2,2
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=5 2+2=4
Calculate the sum for each pair.
a=1 b=4
The solution is the pair that gives sum 5.
\left(k^{2}+k\right)+\left(4k+4\right)
Rewrite k^{2}+5k+4 as \left(k^{2}+k\right)+\left(4k+4\right).
k\left(k+1\right)+4\left(k+1\right)
Factor out k in the first and 4 in the second group.
\left(k+1\right)\left(k+4\right)
Factor out common term k+1 by using distributive property.
k^{2}+5k+4=0
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=\frac{-5±\sqrt{5^{2}-4\times 4}}{2}
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{25-4\times 4}}{2}
Square 5.
k=\frac{-5±\sqrt{25-16}}{2}
Multiply -4 times 4.
k=\frac{-5±\sqrt{9}}{2}
Add 25 to -16.
k=\frac{-5±3}{2}
Take the square root of 9.
k=-\frac{2}{2}
Now solve the equation k=\frac{-5±3}{2} when ± is plus. Add -5 to 3.
k=-1
Divide -2 by 2.
k=-\frac{8}{2}
Now solve the equation k=\frac{-5±3}{2} when ± is minus. Subtract 3 from -5.
k=-4
Divide -8 by 2.
k^{2}+5k+4=\left(k-\left(-1\right)\right)\left(k-\left(-4\right)\right)
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+1\right)\left(k+4\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +5x +4 = 0
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.
r + s = -5 rs = 4
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 = -\frac{5}{2} - u s = -\frac{5}{2} + u
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>
(-\frac{5}{2} - u) (-\frac{5}{2} + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
\frac{25}{4} - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-\frac{25}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{5}{2} - \frac{3}{2} = -4 s = -\frac{5}{2} + \frac{3}{2} = -1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.