Factor the polynomial using special product formulas.25x2−110x+12125x^2-110x+12125x2−110x+121

( 5 x − 11 ) 2 \left(5x-11\right)^2 ( 5 x − 11 ) 2

Factor the polynomial using special product formulas. 25 x 2 − 110 x + 121 25x^2-110x+121 25 x 2 − 110 x + 121

Welcome back, everyone. In this practice problem, we're gonna factor this polynomial over here, this 25x2-110x plus 121 by using special product formulas. And remember the way this works here is I'm gonna take a look at this polynomial, and if I can't factor it by pulling out a greatest common factor or if I can't group it like we can't here, then we're gonna have to use another method. And basically, what happens is if you notice that there's, like, perfect squares or cubes or some of these numbers, you're just gonna try to match these things to one of these equations over here on the right side. So you match your polynomial to the right side over here. And basically, you know, it broke down to whether you have 2 terms or 3 terms because if you can identify that it fits this pattern, then we can just use our special product formulas, and you'll know what the factors are. Okay? So basically, I'm gonna look at this polynomial. I notice that it's 3 terms, and I only have one special product formula with 3 terms, and it's this one over here. So the whole thing is, can I try to and it can I can this I can identify this polynomial as being the form a squared minus 2ab plus b squared? Alright. So let's take a look here. So is my first term, the 25x squared, is that a perfect square? And actually, it is. If this is, if a is equal to 5x, then that means that a squared is 25x squared, and this thing is a perfect square. What about the b term? Is this b term a perfect square? Well, a 121 actually is a perfect square. It's actually just 11 squared. So this b term is equal to 11. So now we have what we have to do is these two terms work out, but what about the middle one? This thing only works if the middle one actually does fit the form 2ab. So using these two things over here, does 2 times a times b actually equal a 100 and negative 110x? Let's find out. 2 times a would be 5x, and then times b would be times 11. So 2 times 5 is 10 x, and then 10 and 10 x times 11 doesn't does actually, in fact, get you to a 110 x. So that means that we're basically done here. We know that this thing is gonna factor into whatever this on the whatever is on the right side over here. Here, we have a squared minus 2ab plus b squared. Remember how this equation works? Whatever sign this is, is the sign of the first, is the sign of the thing between the a squared and the 2ab. So what this actually works out to is if you have a minus b squared, this works out to a squared minus 2ab. So this just becomes, this really just becomes a minus b squared. Alright? That's what our factors are gonna be. So I'm gonna take this expression over here, this my original 25x minus whatever, and this really just breaks down to what a minus b squared is. What is a and b? A is just 5x and b is 11. So this actually just turns out to be 5x minus 11, and that's squared. So this is my answer. If you go ahead and foil this out or multiply it out, you should get back to your original expression. So this is your final answer. Alright. Hopefully, that made sense. Thanks for watching.

Factor the polynomial using special product formulas. 25 x 2 − 110 x + 121 25x^2-110x+121 25 x 2 − 110 x + 121

( 5 x − 11 ) 2 \left(5x-11\right)^2 ( 5 x − 11 ) 2

( 5 x − 10 ) 2 \left(5x-10\right)^2 ( 5 x − 10 ) 2

( 5 x + 11 ) ( 5 x − 11 ) \left(5x+11\right)\left(5x-11\right) ( 5 x + 11 ) ( 5 x − 11 )

( 5 x + 11 ) 2 \left(5x+11\right)^2 ( 5 x + 11 ) 2

0. Fundamental Concepts of Algebra

Factoring Polynomials

Precalculus

0. Fundamental Concepts of Algebra

Factoring Polynomials

Struggling with Precalculus?

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Multiple Choice

Factor the polynomial using special product formulas.
$25x^2-110x+121$

$\left(5x-10\right)^2$

$\left(5x+11\right)\left(5x-11\right)$

$\left(5x+11\right)^2$

$\left(5x-11\right)^2$

Verified step by step guidance

Recognize that the polynomial is a quadratic expression in the form of $ ax^2 + bx + c $. Here, $ a = 25 $, $ b = -110 $, and $ c = 121 $.

Identify that the expression can potentially be factored using the perfect square trinomial formula, which is $ (ax + b)^2 = a^2x^2 + 2abx + b^2 $.

Check if the expression fits the perfect square trinomial pattern by comparing $ 25x^2 - 110x + 121 $ with $ (5x - 11)^2 = 25x^2 - 2 \times 5 \times 11x + 11^2 $.

Calculate $ 2 \times 5 \times 11 = 110 $ and $ 11^2 = 121 $, confirming that the middle term and constant term match the original polynomial.

Conclude that the polynomial $ 25x^2 - 110x + 121 $ can be factored as $ (5x - 11)^2 $, using the special product formula for perfect square trinomials.

7:30m

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