Question #bd0e0

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Question #bd0e0

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Answer 1 · 2017-11-15T03:12:24

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) ${(a + b)}^{2}$ $(a+b)^2$

or

2) ${(a - b)}^{2}$ $(a-b)^2$

If we expand both of those out, we get

1) ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

and

2) ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$

Let's see if $x^{2} - 7 x + 49$ $x^2-7x+49$ 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:

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

$\sqrt{x^{2}} = x$ $sqrt(x^2)=x$ and $\sqrt{49} = \pm 7$ $sqrt(49)=+-7$

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

$(2) (x) (+ 7) = 14 x$ $(2)(x)(+7)=14x$

Nope. Let's try the other combination:

$(2) (x) (- 7) = - 14 x$ $(2)(x)(-7)=-14x$

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

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