Factor the polynomial using special product formulas. x 2 49 − 9 \frac{x^2}{49}-9 49 x 2 − 9

Alright, everyone. Let's keep going with some more practice here. We're gonna factor this polynomial by using our special product formulas. Let's go ahead and get to it. So can we identify can we identify this polynomial as being one of these patterns over here? A2b2a3b3 or plus b cubed, and then we also have this trinomial over here. Well, if you take a look, I've got one number, I have one term, and then I've got minus another term. So really, it can be one of 2 possibilities. It can be either this one or this one. So let's take a look here. Is x squared over 49? Is that a perfect square like over here? Or is it a perfect cube? Well, x squared is a perfect square. It can't be a perfect cube. And what I also notice here is that I have 9, which is a perfect square. That's not a perfect cube. So it looks like what this actually becomes is it's not gonna become this one. It looks like it does fit this pattern over here of a squared minus b squared. Now what that means here is that my factors on the on the on the other side are really just gonna be, a plus b, a minus b. This is just a difference of squares. All I have to do now is just identify a perfect square of something. So what is a? Well, if a is x squared of 49, remember fractions, when you square them, you just square the top and the bottom. So what is the x squared? Well, that's just x on top. What about the 49? Is that a perfect square? Well, 49 is a perfect square of 7. So if I had a, which is equal to x over 7, x divided by 7, then that means if I square this thing, I should just get back to x squared minus 40 or over 49. So it looks like a is equal to x divided by 7. What about the b term? Well, the b term, if b squared is just 9, that means that b is equal to 3. So that means that I'm basically done here. What this factors to is it's gonna be a+b and a minus b. So what's my a? It's a it's x divided by 7. So it's going to be x divided by 7 plus 3, and then x divided by 7 minus 3. If you take this expression and you foil it out, what's gonna happen is then the middle two terms will cancel. You're just gonna get x over 7 squared, which gets you back to x squared over 49, and the 3 and the 3 just become 9. So that's how to factor this kind of expression. Hopefully that made sense, and, thanks for watching.

Factor the polynomial using special product formulas. x 2 49 − 9 \frac{x^2}{49}-9 49 x 2 − 9

( x 7 − 3 ) 2 \left(\frac{x}{7}-3\right)^2 ( 7 x − 3 ) 2

( x 7 + 3 ) ( x 7 − 3 ) \left(\frac{x}{7}+3\right)\left(\frac{x}{7}-3\right) ( 7 x + 3 ) ( 7 x − 3 )

( 7 x − 3 ) 2 \left(7x-3\right)^2 ( 7 x − 3 ) 2

( 7 x + 3 ) ( 7 x − 3 ) \left(7x+3\right)\left(7x-3\right) ( 7 x + 3 ) ( 7 x − 3 )

0. Fundamental Concepts of Algebra

Factoring Polynomials

Precalculus

0. Fundamental Concepts of Algebra

Factoring Polynomials

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Multiple Choice

Factor the polynomial using special product formulas.
$\frac{x^2}{49}-9$

$\left(\frac{x}{7}-3\right)^2$

$\left(\frac{x}{7}+3\right)\left(\frac{x}{7}-3\right)$

$\left(7x-3\right)^2$

$\left(7x+3\right)\left(7x-3\right)$

Verified step by step guidance

Recognize that the given expression $ \frac{x^2}{49} - 9 $ is a difference of squares. The difference of squares formula is $ a^2 - b^2 = (a + b)(a - b) $.

Identify $ a $ and $ b $ in the expression $ \frac{x^2}{49} - 9 $. Here, $ a = \frac{x}{7} $ and $ b = 3 $, since $ \left(\frac{x}{7}\right)^2 = \frac{x^2}{49} $ and $ 3^2 = 9 $.

Apply the difference of squares formula: $ \left(\frac{x}{7}\right)^2 - 3^2 = \left(\frac{x}{7} + 3\right)\left(\frac{x}{7} - 3\right) $.

Notice that the expression can be rewritten as $ \left(\frac{x}{7} - 3\right)^2 $, which is a perfect square trinomial.

Factor the perfect square trinomial using the formula $ (a - b)^2 = (a - b)(a - b) $, resulting in $ \left(\frac{x}{7} - 3\right)\left(\frac{x}{7} - 3\right) $.