Factor
\left(x+1\right)\left(x+5\right)
Evaluate
\left(x+1\right)\left(x+5\right)
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a+b=6 ab=1\times 5=5
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
a=1 b=5
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(x^{2}+x\right)+\left(5x+5\right)
Rewrite x^{2}+6x+5 as \left(x^{2}+x\right)+\left(5x+5\right).
x\left(x+1\right)+5\left(x+1\right)
Factor out x in the first and 5 in the second group.
\left(x+1\right)\left(x+5\right)
Factor out common term x+1 by using distributive property.
x^{2}+6x+5=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.
x=\frac{-6±\sqrt{6^{2}-4\times 5}}{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.
x=\frac{-6±\sqrt{36-4\times 5}}{2}
Square 6.
x=\frac{-6±\sqrt{36-20}}{2}
Multiply -4 times 5.
x=\frac{-6±\sqrt{16}}{2}
Add 36 to -20.
x=\frac{-6±4}{2}
Take the square root of 16.
x=-\frac{2}{2}
Now solve the equation x=\frac{-6±4}{2} when ± is plus. Add -6 to 4.
x=-1
Divide -2 by 2.
x=-\frac{10}{2}
Now solve the equation x=\frac{-6±4}{2} when ± is minus. Subtract 4 from -6.
x=-5
Divide -10 by 2.
x^{2}+6x+5=\left(x-\left(-1\right)\right)\left(x-\left(-5\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 -5 for x_{2}.
x^{2}+6x+5=\left(x+1\right)\left(x+5\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +6x +5 = 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 = -6 rs = 5
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 = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>
(-3 - u) (-3 + u) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
9 - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-9 = -4
Simplify the expression by subtracting 9 on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - 2 = -5 s = -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.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}