Solve 10-3a-a^2

$$10 - 3 a - a ^ { 2 }$$

Factor

$$\left(2-a\right)\left(a+5\right)$$

Solution Steps

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

$$-a^{2}-3a+10$$

Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa+10. To find p and q, set up a system to be solved.

$$p+q=-3 pq=-10=-10$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$p=2 q=-5$$

Rewrite -a^{2}-3a+10 as \left(-a^{2}+2a\right)+\left(-5a+10\right).

$$\left(-a^{2}+2a\right)+\left(-5a+10\right)$$

Factor out a in the first and 5 in the second group.

$$a\left(-a+2\right)+5\left(-a+2\right)$$

Factor out common term -a+2 by using distributive property.

$$\left(-a+2\right)\left(a+5\right)$$

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.

$$-a^{2}-3a+10=0$$

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.

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-\left(-3\right)±\sqrt{9+4\times 10}}{2\left(-1\right)}$$

Multiply 4 times 10.

$$a=\frac{-\left(-3\right)±\sqrt{9+40}}{2\left(-1\right)}$$

Add 9 to 40.

$$a=\frac{-\left(-3\right)±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$a=\frac{-\left(-3\right)±7}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$a=\frac{3±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{3±7}{-2}$$

Now solve the equation a=\frac{3±7}{-2} when ± is plus. Add 3 to 7.

$$a=\frac{10}{-2}$$

Divide 10 by -2.

$$a=-5$$

Now solve the equation a=\frac{3±7}{-2} when ± is minus. Subtract 7 from 3.

$$a=-\frac{4}{-2}$$

Divide -4 by -2.

$$a=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 2 for x_{2}.

$$-a^{2}-3a+10=-\left(a-\left(-5\right)\right)\left(a-2\right)$$

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

$$-a^{2}-3a+10=-\left(a+5\right)\left(a-2\right)$$

Evaluate

$$\left(2-a\right)\left(a+5\right)$$