Solve 3x^2-6x-9

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

Factor

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

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.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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.

$$a=-3 b=1$$

Rewrite x^{2}-2x-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.

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

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.

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

Square -6.

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

Multiply -4 times 3.

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

Multiply -12 times -9.

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

Add 36 to 108.

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

Take the square root of 144.

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

The opposite of -6 is 6.

$$x=\frac{6±12}{2\times 3}$$

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}.

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

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

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

Evaluate

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