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a+b=5 ab=2\times 2=4
Factor the expression by grouping. First, the expression needs to be rewritten as 2p^{2}+ap+bp+2. 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(2p^{2}+p\right)+\left(4p+2\right)
Rewrite 2p^{2}+5p+2 as \left(2p^{2}+p\right)+\left(4p+2\right).
p\left(2p+1\right)+2\left(2p+1\right)
Factor out p in the first and 2 in the second group.
\left(2p+1\right)\left(p+2\right)
Factor out common term 2p+1 by using distributive property.
2p^{2}+5p+2=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{-5±\sqrt{5^{2}-4\times 2\times 2}}{2\times 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{-5±\sqrt{25-4\times 2\times 2}}{2\times 2}
Square 5.
p=\frac{-5±\sqrt{25-8\times 2}}{2\times 2}
Multiply -4 times 2.
p=\frac{-5±\sqrt{25-16}}{2\times 2}
Multiply -8 times 2.
p=\frac{-5±\sqrt{9}}{2\times 2}
Add 25 to -16.
p=\frac{-5±3}{2\times 2}
Take the square root of 9.
p=\frac{-5±3}{4}
Multiply 2 times 2.
p=-\frac{2}{4}
Now solve the equation p=\frac{-5±3}{4} when ± is plus. Add -5 to 3.
p=-\frac{1}{2}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
p=-\frac{8}{4}
Now solve the equation p=\frac{-5±3}{4} when ± is minus. Subtract 3 from -5.
p=-2
Divide -8 by 4.
2p^{2}+5p+2=2\left(p-\left(-\frac{1}{2}\right)\right)\left(p-\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 -\frac{1}{2} for x_{1} and -2 for x_{2}.
2p^{2}+5p+2=2\left(p+\frac{1}{2}\right)\left(p+2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
2p^{2}+5p+2=2\times \frac{2p+1}{2}\left(p+2\right)
Add \frac{1}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
2p^{2}+5p+2=\left(2p+1\right)\left(p+2\right)
Cancel out 2, the greatest common factor in 2 and 2.
x ^ 2 +\frac{5}{2}x +1 = 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.This is achieved by dividing both sides of the equation by 2
r + s = -\frac{5}{2} rs = 1
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}{4} - u s = -\frac{5}{4} + u
Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. 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}{4} - u) (-\frac{5}{4} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{25}{16} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{25}{16} = -\frac{9}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}
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}{4} - \frac{3}{4} = -2 s = -\frac{5}{4} + \frac{3}{4} = -0.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.