Solve -10+a^2-3a

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

Factor

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

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.

$$p+q=-3 pq=1\left(-10\right)=-10$$

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.

$$p=-5 q=2$$

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

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

Factor out a in the first and 2 in the second group.

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

Factor out common term a-5 by using distributive property.

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

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.

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

Square -3.

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

Multiply -4 times -10.

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

Add 9 to 40.

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

Take the square root of 49.

$$a=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$a=\frac{3±7}{2}$$

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}.

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

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

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