# How do you factor completely x^2+ 4x + 2?

Nov 14, 2015

Not sure about my solution. Have a look!!

#### Explanation:

Certainly at the moment I can not think of an alternative to:

Factors of ${x}^{2}$ can only be $\pm \left(x \text{ and } x\right)$
Factors of 2 can only be $\pm \left(1 \text{ and } 2\right)$. You could have $\sqrt{2} \times \sqrt{2} \textcolor{w h i t e}{x}$ I suppose!

The problem is that $\left(x \times 1\right) + \left(x \times 2\right) \ne 4 x$

Suppose we had instead $\left(x \times 2\right) + \left(x \times 2\right) = 4 x$
This works for the coefficient (number in front) of $x$ but fails for the constant of 2 at the end of the expression.

$2 \times 2 = 4$ this is 2 too much. We can get over this by subtracting the amount that is too much.

So
${x}^{2} + 4 x + 2 = \left(x + 2\right) \left(x + 2\right) - 2 = {\left(x + 2\right)}^{2} - 2$

The problem is that when asked to $\textcolor{g r e e n}{\text{factorise completely}}$ I expect that I am able to do so. Consequently, $\textcolor{b l u e}{\text{either I have gone wrong or the question is not quite correct.}}$

Nov 14, 2015

Complete the square, then use the difference of squares identity to find:

${x}^{2} + 4 x + 2 = \left(x + 2 - \sqrt{2}\right) \left(x + 2 + \sqrt{2}\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

In our example:

${x}^{2} + 4 x + 2$

$= {x}^{2} + 4 x + 4 - 2$

$= {\left(x + 2\right)}^{2} - 2$

$= {\left(x + 2\right)}^{2} - {\left(\sqrt{2}\right)}^{2}$

$= \left(x + 2 - \sqrt{2}\right) \left(x + 2 + \sqrt{2}\right)$

In general, given $a {x}^{2} + b x + c$, you can express it in terms of a square and a constant term as follows:

$a {x}^{2} + b x + c$

$= a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

$= a \left({\left(x + \frac{b}{2 a}\right)}^{2} - \left({b}^{2} / \left(4 {a}^{2}\right) - \frac{c}{a}\right)\right)$

$= a \left({\left(x + \frac{b}{2 a}\right)}^{2} - {\left(\sqrt{{b}^{2} / \left(4 {a}^{2}\right) - \frac{c}{a}}\right)}^{2}\right)$

$= a \left(x + \frac{b}{2 a} - \sqrt{{b}^{2} / \left(4 {a}^{2}\right) - \frac{c}{a}}\right) \left(x + \frac{b}{2 a} + \sqrt{{b}^{2} / \left(4 {a}^{2}\right) - \frac{c}{a}}\right)$

$= a \left(x + \frac{b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x + \frac{b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$