Solve 3x^2-6x+0

$$3 x ^ { 2 } - 6 x + 0$$

Evaluate

$$3x\left(x-2\right)$$

Solution Steps

Anything plus zero gives itself.

$$3x^{2}-6x$$

Steps Using Direct Factoring Method

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 3

$$x ^ 2 -2x +0 = 0$$

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

$$r + s = 2 rs = 0$$

Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 0

$$(1 - u) (1 + u) = 0$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$1 - u^2 = 0$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = 0-1 = -1$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =1 - 1 = 0 s = 1 + 1 = 2$$

Factor

$$3x\left(x-2\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}-2x\right)$$

Consider x^{2}-2x. Factor out x.

$$x\left(x-2\right)$$

Rewrite the complete factored expression.

$$3x\left(x-2\right)$$

Steps Using the Quadratic Formula

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.

$$3x^{2}-6x=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\times 3}$$

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2\times 3}$$

The opposite of -6 is 6.

$$x=\frac{6±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±6}{6}$$

Now solve the equation x=\frac{6±6}{6} when ± is plus. Add 6 to 6.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{6±6}{6} when ± is minus. Subtract 6 from 6.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and 0 for x_{2}.

$$3x^{2}-6x=3\left(x-2\right)x$$