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a+b=3 ab=1\times 2=2
Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+2. To find a and b, set up a system to be solved.
a=1 b=2
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(y^{2}+y\right)+\left(2y+2\right)
Rewrite y^{2}+3y+2 as \left(y^{2}+y\right)+\left(2y+2\right).
y\left(y+1\right)+2\left(y+1\right)
Factor out y in the first and 2 in the second group.
\left(y+1\right)\left(y+2\right)
Factor out common term y+1 by using distributive property.
y^{2}+3y+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.
y=\frac{-3±\sqrt{3^{2}-4\times 2}}{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.
y=\frac{-3±\sqrt{9-4\times 2}}{2}
Square 3.
y=\frac{-3±\sqrt{9-8}}{2}
Multiply -4 times 2.
y=\frac{-3±\sqrt{1}}{2}
Add 9 to -8.
y=\frac{-3±1}{2}
Take the square root of 1.
y=-\frac{2}{2}
Now solve the equation y=\frac{-3±1}{2} when ± is plus. Add -3 to 1.
y=-1
Divide -2 by 2.
y=-\frac{4}{2}
Now solve the equation y=\frac{-3±1}{2} when ± is minus. Subtract 1 from -3.
y=-2
Divide -4 by 2.
y^{2}+3y+2=\left(y-\left(-1\right)\right)\left(y-\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 -1 for x_{1} and -2 for x_{2}.
y^{2}+3y+2=\left(y+1\right)\left(y+2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +3x +2 = 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 = -3 rs = 2
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{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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{3}{2} - u) (-\frac{3}{2} + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
\frac{9}{4} - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-\frac{9}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{9}{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{3}{2} - \frac{1}{2} = -2 s = -\frac{3}{2} + \frac{1}{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.