Solve -3x^2+12x-9

$$- 3 x ^ { 2 } + 12 x - 9$$

Factor

$$3\left(1-x\right)\left(x-3\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}+4x-3\right)$$

Consider -x^{2}+4x-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.

$$a+b=4 ab=-\left(-3\right)=3$$

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.

$$a=3 b=1$$

Rewrite -x^{2}+4x-3 as \left(-x^{2}+3x\right)+\left(x-3\right).

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

Factor out -x in -x^{2}+3x.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+1\right)$$

Rewrite the complete factored expression.

$$3\left(x-3\right)\left(-x+1\right)$$

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.

$$-3x^{2}+12x-9=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-12±\sqrt{144+12\left(-9\right)}}{2\left(-3\right)}$$

Multiply 12 times -9.

$$x=\frac{-12±\sqrt{144-108}}{2\left(-3\right)}$$

Add 144 to -108.

$$x=\frac{-12±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{-12±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-12±6}{-6}$$

Now solve the equation x=\frac{-12±6}{-6} when ± is plus. Add -12 to 6.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-12±6}{-6} when ± is minus. Subtract 6 from -12.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

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}.

$$-3x^{2}+12x-9=-3\left(x-1\right)\left(x-3\right)$$

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.

$$x ^ 2 -4x +3 = 0$$

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

$$r + s = 4 rs = 3$$

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>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(2 - u) (2 + u) = 3$$

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

$$4 - u^2 = 3$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 3-4 = -1$$

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.

$$r =2 - 1 = 1 s = 2 + 1 = 3$$

Evaluate

$$3\left(1-x\right)\left(x-3\right)$$