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a+b=4 ab=1\times 3=3
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.
a=1 b=3
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(3x+3\right)
Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).
x\left(x+1\right)+3\left(x+1\right)
Factor out x in the first and 3 in the second group.
\left(x+1\right)\left(x+3\right)
Factor out common term x+1 by using distributive property.
x^{2}+4x+3=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{-4±\sqrt{4^{2}-4\times 3}}{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{-4±\sqrt{16-4\times 3}}{2}
Square 4.
x=\frac{-4±\sqrt{16-12}}{2}
Multiply -4 times 3.
x=\frac{-4±\sqrt{4}}{2}
Add 16 to -12.
x=\frac{-4±2}{2}
Take the square root of 4.
x=-\frac{2}{2}
Now solve the equation x=\frac{-4±2}{2} when ± is plus. Add -4 to 2.
x=-1
Divide -2 by 2.
x=-\frac{6}{2}
Now solve the equation x=\frac{-4±2}{2} when ± is minus. Subtract 2 from -4.
x=-3
Divide -6 by 2.
x^{2}+4x+3=\left(x-\left(-1\right)\right)\left(x-\left(-3\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 -3 for x_{2}.
x^{2}+4x+3=\left(x+1\right)\left(x+3\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +4x +3 = 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 = -4 rs = 3
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 = -2 - u s = -2 + u
Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>
(-2 - u) (-2 + u) = 3
To solve for unknown quantity u, substitute these in the product equation rs = 3
4 - u^2 = 3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 3-4 = -1
Simplify the expression by subtracting 4 on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-2 - 1 = -3 s = -2 + 1 = -1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.