Factor
7\left(x^{2}-4x+5\right)
Evaluate
7\left(x^{2}-4x+5\right)
Graph
Share
Copied to clipboard
7\left(x^{2}-4x+5\right)
Factor out 7. Polynomial x^{2}-4x+5 is not factored since it does not have any rational roots.
7x^{2}-28x+35=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{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 7\times 35}}{2\times 7}
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{-\left(-28\right)±\sqrt{784-4\times 7\times 35}}{2\times 7}
Square -28.
x=\frac{-\left(-28\right)±\sqrt{784-28\times 35}}{2\times 7}
Multiply -4 times 7.
x=\frac{-\left(-28\right)±\sqrt{784-980}}{2\times 7}
Multiply -28 times 35.
x=\frac{-\left(-28\right)±\sqrt{-196}}{2\times 7}
Add 784 to -980.
7x^{2}-28x+35
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 -4x +5 = 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 7
r + s = 4 rs = 5
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) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
4 - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-4 = 1
Simplify the expression by subtracting 4 on both sides
u^2 = -1 u = \pm\sqrt{-1} = \pm i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =2 - i s = 2 + i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}