Skip to main content
Factor
Tick mark Image
Evaluate
Tick mark Image
Graph

Similar Problems from Web Search

Share

2\left(2x^{2}+2x+1\right)
Factor out 2. Polynomial 2x^{2}+2x+1 is not factored since it does not have any rational roots.
4x^{2}+4x+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.
x=\frac{-4±\sqrt{4^{2}-4\times 4\times 2}}{2\times 4}
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 4\times 2}}{2\times 4}
Square 4.
x=\frac{-4±\sqrt{16-16\times 2}}{2\times 4}
Multiply -4 times 4.
x=\frac{-4±\sqrt{16-32}}{2\times 4}
Multiply -16 times 2.
x=\frac{-4±\sqrt{-16}}{2\times 4}
Add 16 to -32.
4x^{2}+4x+2
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
x ^ 2 +1x +\frac{1}{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.This is achieved by dividing both sides of the equation by 4
r + s = -1 rs = \frac{1}{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{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) = \frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}
\frac{1}{4} - u^2 = \frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{1}{4} u = \pm\sqrt{-\frac{1}{4}} = \pm \frac{1}{2}i
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{1}{2}i = -0.500 - 0.500i s = -\frac{1}{2} + \frac{1}{2}i = -0.500 + 0.500i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.