Solve -3x^2-6x+9

$$- 3 x ^ { 2 } - 6 x + 9$$

Factor

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

Solution Steps

Factor out 3.

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

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

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

Factor out x in the first and 3 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$-3x^{2}-6x+9=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+12\times 9}}{2\left(-3\right)}$$

Multiply 12 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36+108}}{2\left(-3\right)}$$

Add 36 to 108.

$$x=\frac{-\left(-6\right)±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-6\right)±12}{2\left(-3\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{6±12}{-6}$$

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

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

Divide 18 by -6.

$$x=-3$$

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

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

Divide -6 by -6.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and 1 for x_{2}.

$$-3x^{2}-6x+9=-3\left(x-\left(-3\right)\right)\left(x-1\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}-6x+9=-3\left(x+3\right)\left(x-1\right)$$

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.

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

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

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

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 = -3

$$(-1 - u) (-1 + u) = -3$$

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

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

Simplify the expression by subtracting 1 on both sides

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

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

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

Evaluate

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