# Question #bd0e0

Nov 15, 2017

You can't. This expression is unfactorable.

#### Explanation:

A perfect square trinomial is a trinomial (a polynomial with three terms) that can be factored down into one of two forms:

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

or

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

If we expand both of those out, we get

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

and

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

Let's see if ${x}^{2} - 7 x + 49$ is a perfect square trinomial. According to the expansion, we see that the first and last term have to be perfect squares. Let's check that:

${x}^{2}$ and $49$ are the first and last term, respectively.

$\sqrt{{x}^{2}} = x$ and $\sqrt{49} = \pm 7$

The two terms are perfect squares! Now let's see - the middle term has to be twice the product of the two squares.

Our two squares are $x$ and $\pm 7$ (NOTE: We have two different combinations for $\pm 7$: $+ 7$ and $- 7$, so we have to test both of them to see if they fit the properties of the middle term of a perfect square term).

$\left(2\right) \left(x\right) \left(+ 7\right) = 14 x$

Nope. Let's try the other combination:

$\left(2\right) \left(x\right) \left(- 7\right) = - 14 x$

Nope either. Even though our first and last terms were perfect squares, the middle term did not have the properties of a perfect square trinomial.

Therefore, ${x}^{2} - 7 x + 49$ is not a perfect square trinomial. And so, it cannot be expressed as the product of two binomials.