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a+b=-5 ab=1\times 6=6
Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+6. To find a and b, set up a system to be solved.
-1,-6 -2,-3
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.
-1-6=-7 -2-3=-5
Calculate the sum for each pair.
a=-3 b=-2
The solution is the pair that gives sum -5.
\left(p^{2}-3p\right)+\left(-2p+6\right)
Rewrite p^{2}-5p+6 as \left(p^{2}-3p\right)+\left(-2p+6\right).
p\left(p-3\right)-2\left(p-3\right)
Factor out p in the first and -2 in the second group.
\left(p-3\right)\left(p-2\right)
Factor out common term p-3 by using distributive property.
p^{2}-5p+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.
p=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{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.
p=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}
Square -5.
p=\frac{-\left(-5\right)±\sqrt{25-24}}{2}
Multiply -4 times 6.
p=\frac{-\left(-5\right)±\sqrt{1}}{2}
Add 25 to -24.
p=\frac{-\left(-5\right)±1}{2}
Take the square root of 1.
p=\frac{5±1}{2}
The opposite of -5 is 5.
p=\frac{6}{2}
Now solve the equation p=\frac{5±1}{2} when ± is plus. Add 5 to 1.
p=3
Divide 6 by 2.
p=\frac{4}{2}
Now solve the equation p=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.
p=2
Divide 4 by 2.
p^{2}-5p+6=\left(p-3\right)\left(p-2\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}.
x ^ 2 -5x +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 = 5 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{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) = 6
To solve for unknown quantity u, substitute these in the product equation rs = 6
\frac{25}{4} - u^2 = 6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 6-\frac{25}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{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{1}{2} = 2 s = \frac{5}{2} + \frac{1}{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.