Factoring $s^2+4s+13$

Question

I was looking at an example, and it was factored as follow:

$$ s^{2}+4s+13 = (s+2)^{2}+9 $$

How can we do that?

Answer 1

$s^2+4s+13=s^2+4s+4+9=(s+2)^2+9$

Answer 2

Notice that $$s^2+2As+A^2=(s+A)^2.$$ So, in your case, $$s^2+4s+13=(s^2+2\times 2s+2^2)-2^2+13=(s+2)^2+9.$$

Answer 3

By the binomial formula you have

$$(a+b)^2 = a^2 + 2ab + b^2$$

In you example $a$ is equal to $s$ and $b$ is equal to $2$, so just add $2^2$ and substrct it again

$$s^2 + 4a + 13 = s^2 + 2\cdot 2a + 2^2 -2^2+13$$

Now apply the binomial formula and you will get

$$ (s+2)^2-2^2+13 = (s+2)^2 + 9 $$

Answer 4

By "completing the square"

$s^2+4s+13=s^2+4s+4+9=(s+2)^2+9$

Answer 5

By using the method of Completing the Square: $$s^2+4s+13=0\implies s^2+4s=-13$$

Divide the $b$ term by $2$ and square this, then add to both sides:

$$\underbrace{s^2 +4s + \left(\frac{4}{2}\right)^2}_{(s+2)^2}=-13+\left(\frac{4}{2}\right)^2$$

The LHS becomes (after some calculation): $$-13+\left(\frac{4}{2}\right)^2=-\frac{52}{4}+\frac{16}{4}=-\frac{36}{4}=-9$$

Therefore we have $$(s+2)^2=-9\implies (s+2)^2+9$$ So $$s^2+4s+13=(s+2)^2+9$$

Answer 6

This is called completing the square. We can write any quadratic in the form $a(s-h)^2 + k$. If the quadratic is monic (i.e. the coefficient of the squared term is one), then it is of the form $(s - h)^2 + k$.

If we want to write the quadratic $s^2 + 4s + 13$ in the form $(s-h)^2 + k$, we expand the latter to obtain $s^2 - 2hs + h^2 + k$, and then compare coefficients. Doing so yields $2h = 4$, so $h = -2$, and $h^2 + k = 13$, from which it follows $k = 9$. Therefore $s^2 + 4s + 13 = (s+2)^2 + 9$.

If you try this for other monic quadratics, you will see the same steps always occur, which is why completing the square is often taught only as a method (which is why some people don't understand it). The steps for a monic quadratic $s^2 + bs + c$ are as follows (if the quadratic is not monic, first factor out the coefficient of the squared term):

1. Halve the coefficient of $s$ (i.e. $\frac{b}{2}$), this is your $-h$ (for the exact same reason as above).
2. Square this and subtract it from the constant term (i.e. $c - \frac{b^2}{4}$), this is your $k$ (for the exact same reason as above).
3. We have $s^2 + bs + c = \left(s + \frac{b}{2}\right)^2 + \left(c-\frac{b^2}{4}\right)$.

Note, some people write this process out as

\begin{align*} s^2 + bs + c &= s^2 + bs + \frac{b^2}{4} + c - \frac{b^2}{4}\\ &=s^2 + 2\frac{b}{2}s + \left(\frac{b}{2}\right)^2 + c - \frac{b^2}{4}\\ & = \left(s + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right). \end{align*}