Factor
\left(m-3\right)\left(m+2\right)
Evaluate
\left(m-3\right)\left(m+2\right)
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a+b=-1 ab=1\left(-6\right)=-6
Factor the expression by grouping. First, the expression needs to be rewritten as m^{2}+am+bm-6. To find a and b, set up a system to be solved.
1,-6 2,-3
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. List all such integer pairs that give product -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(m^{2}-3m\right)+\left(2m-6\right)
Rewrite m^{2}-m-6 as \left(m^{2}-3m\right)+\left(2m-6\right).
m\left(m-3\right)+2\left(m-3\right)
Factor out m in the first and 2 in the second group.
\left(m-3\right)\left(m+2\right)
Factor out common term m-3 by using distributive property.
m^{2}-m-6=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.
m=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{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.
m=\frac{-\left(-1\right)±\sqrt{1+24}}{2}
Multiply -4 times -6.
m=\frac{-\left(-1\right)±\sqrt{25}}{2}
Add 1 to 24.
m=\frac{-\left(-1\right)±5}{2}
Take the square root of 25.
m=\frac{1±5}{2}
The opposite of -1 is 1.
m=\frac{6}{2}
Now solve the equation m=\frac{1±5}{2} when ± is plus. Add 1 to 5.
m=3
Divide 6 by 2.
m=-\frac{4}{2}
Now solve the equation m=\frac{1±5}{2} when ± is minus. Subtract 5 from 1.
m=-2
Divide -4 by 2.
m^{2}-m-6=\left(m-3\right)\left(m-\left(-2\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -2 for x_{2}.
m^{2}-m-6=\left(m-3\right)\left(m+2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 -1x -6 = 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 = 1 rs = -6
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{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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{1}{2} - u) (\frac{1}{2} + u) = -6
To solve for unknown quantity u, substitute these in the product equation rs = -6
\frac{1}{4} - u^2 = -6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -6-\frac{1}{4} = -\frac{25}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{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{1}{2} - \frac{5}{2} = -2 s = \frac{1}{2} + \frac{5}{2} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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y = 3x + 4
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Matrix
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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}