Solve -20a-100-a^2=

$$- 20 a - 100 - a ^ { 2 } =$$

Factor

$$-\left(a+10\right)^{2}$$

Solution Steps

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

$$-a^{2}-20a-100$$

Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-100. To find p and q, set up a system to be solved.

$$p+q=-20 pq=-\left(-100\right)=100$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$p=-10 q=-10$$

Rewrite -a^{2}-20a-100 as \left(-a^{2}-10a\right)+\left(-10a-100\right).

$$\left(-a^{2}-10a\right)+\left(-10a-100\right)$$

Factor out -a in the first and -10 in the second group.

$$-a\left(a+10\right)-10\left(a+10\right)$$

Factor out common term a+10 by using distributive property.

$$\left(a+10\right)\left(-a-10\right)$$

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.

$$-a^{2}-20a-100=0$$

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.

$$a=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-1\right)\left(-100\right)}}{2\left(-1\right)}$$

Square -20.

$$a=\frac{-\left(-20\right)±\sqrt{400-4\left(-1\right)\left(-100\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-\left(-20\right)±\sqrt{400+4\left(-100\right)}}{2\left(-1\right)}$$

Multiply 4 times -100.

$$a=\frac{-\left(-20\right)±\sqrt{400-400}}{2\left(-1\right)}$$

Add 400 to -400.

$$a=\frac{-\left(-20\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$a=\frac{-\left(-20\right)±0}{2\left(-1\right)}$$

The opposite of -20 is 20.

$$a=\frac{20±0}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{20±0}{-2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -10 for x_{1} and -10 for x_{2}.

$$-a^{2}-20a-100=-\left(a-\left(-10\right)\right)\left(a-\left(-10\right)\right)$$

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

$$-a^{2}-20a-100=-\left(a+10\right)\left(a+10\right)$$

Evaluate

$$-\left(a+10\right)^{2}$$