Solve 11a+a^2+28

$$11 a + a ^ { 2 } + 28$$

Factor

$$\left(a+4\right)\left(a+7\right)$$

Solution Steps

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

$$a^{2}+11a+28$$

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+28. To find p and q, set up a system to be solved.

$$p+q=11 pq=1\times 28=28$$

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 28.

$$1,28 2,14 4,7$$

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 11.

$$p=4 q=7$$

Rewrite a^{2}+11a+28 as \left(a^{2}+4a\right)+\left(7a+28\right).

$$\left(a^{2}+4a\right)+\left(7a+28\right)$$

Factor out a in the first and 7 in the second group.

$$a\left(a+4\right)+7\left(a+4\right)$$

Factor out common term a+4 by using distributive property.

$$\left(a+4\right)\left(a+7\right)$$

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.

$$a^{2}+11a+28=0$$

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.

$$a=\frac{-11±\sqrt{11^{2}-4\times 28}}{2}$$

Square 11.

$$a=\frac{-11±\sqrt{121-4\times 28}}{2}$$

Multiply -4 times 28.

$$a=\frac{-11±\sqrt{121-112}}{2}$$

Add 121 to -112.

$$a=\frac{-11±\sqrt{9}}{2}$$

Take the square root of 9.

$$a=\frac{-11±3}{2}$$

Now solve the equation a=\frac{-11±3}{2} when ± is plus. Add -11 to 3.

$$a=-\frac{8}{2}$$

Divide -8 by 2.

$$a=-4$$

Now solve the equation a=\frac{-11±3}{2} when ± is minus. Subtract 3 from -11.

$$a=-\frac{14}{2}$$

Divide -14 by 2.

$$a=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and -7 for x_{2}.

$$a^{2}+11a+28=\left(a-\left(-4\right)\right)\left(a-\left(-7\right)\right)$$

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

$$a^{2}+11a+28=\left(a+4\right)\left(a+7\right)$$

Evaluate

$$\left(a+4\right)\left(a+7\right)$$