Solve 1*a^2+11a+10=

$$1 \cdot a ^ { 2 } + 11 a + 10 =$$

Factor

$$\left(a+1\right)\left(a+10\right)$$

Solution Steps

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.

$$p+q=11 pq=1\times 10=10$$

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 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$p=1 q=10$$

Rewrite a^{2}+11a+10 as \left(a^{2}+a\right)+\left(10a+10\right).

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

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

$$a\left(a+1\right)+10\left(a+1\right)$$

Factor out common term a+1 by using distributive property.

$$\left(a+1\right)\left(a+10\right)$$

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.

$$a^{2}+11a+10=0$$

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.

$$a=\frac{-11±\sqrt{11^{2}-4\times 10}}{2}$$

Square 11.

$$a=\frac{-11±\sqrt{121-4\times 10}}{2}$$

Multiply -4 times 10.

$$a=\frac{-11±\sqrt{121-40}}{2}$$

Add 121 to -40.

$$a=\frac{-11±\sqrt{81}}{2}$$

Take the square root of 81.

$$a=\frac{-11±9}{2}$$

Now solve the equation a=\frac{-11±9}{2} when ± is plus. Add -11 to 9.

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

Divide -2 by 2.

$$a=-1$$

Now solve the equation a=\frac{-11±9}{2} when ± is minus. Subtract 9 from -11.

$$a=-\frac{20}{2}$$

Divide -20 by 2.

$$a=-10$$

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

$$a^{2}+11a+10=\left(a-\left(-1\right)\right)\left(a-\left(-10\right)\right)$$

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

$$a^{2}+11a+10=\left(a+1\right)\left(a+10\right)$$

Evaluate

$$\left(a+1\right)\left(a+10\right)$$