Solve 6x-x^2-9=

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -x^{2}+6x-9 as \left(-x^{2}+3x\right)+\left(3x-9\right).

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

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

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

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

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

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.

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

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-1\right)}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=\frac{-6±0}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±0}{-2}$$

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 3 for x_{2}.

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

Evaluate

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