Solve 6a-a^2-9

$$6 a - a ^ { 2 } - 9$$

Factor

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

Solution Steps

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

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

Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-9. To find p and q, set up a system to be solved.

$$p+q=6 pq=-\left(-9\right)=9$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q 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.

$$p=3 q=3$$

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

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

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

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

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

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

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.

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

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.

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

Square 6.

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

Multiply -4 times -1.

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

Multiply 4 times -9.

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

Add 36 to -36.

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

Take the square root of 0.

$$a=\frac{-6±0}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-6±0}{-2}$$

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}.

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

Evaluate

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