# How do you determine whether 9x^2+6x+1 is a perfect square trinomial?

Nov 22, 2016

$9 {x}^{2} + 6 x + 1 = {\left(3 x + 1\right)}^{2}$

#### Explanation:

Note that in general:

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

Given:

$9 {x}^{2} + 6 x + 1$

Note that $9 {x}^{2} = {\left(3 x\right)}^{2}$ and $1 = {1}^{2}$

So with $a = 3 x$ and $b = 1$ does the middle term match $2 a b$?

$2 \cdot 3 x \cdot 1 = 6 x \text{ }$ - yes

$9 {x}^{2} + 6 x + 1 = {\left(3 x + 1\right)}^{2}$

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Footnote

If you are familiar with square numbers, then you may recognise:

$961 = {31}^{2}$

Notice that this corresponds to:

$9 {x}^{2} + 6 x + 1 = {\left(3 x + 1\right)}^{2}$

This correspondence is no coincidence. Think about putting $x = 10$.

Since the numbers $3$ and $1$ are small enough, there are no carried digits, so the pattern of digits in the product corresponds.

This can help in general with quickly spotting some perfect square trinomials and other binomial products with small coefficients:

$4 {x}^{2} + 4 x + 1 = {\left(2 x + 1\right)}^{2} \text{ }$ like $\text{ } 441 = {21}^{2}$

${x}^{3} + 3 {x}^{2} + 3 x + 1 = {\left(x + 1\right)}^{3} \text{ }$ like $\text{ } 1331 = {11}^{3}$