Square each binomial.(x2+1)2(x^2+1)^2

x 4 + 2 x 2 + 1 x^4+2x^2+1 x 4 + 2 x 2 + 1

Square each binomial. ( x 2 + 1 ) 2 (x^2+1)^2

Folks. So, let's take a look at our next practice problem here. Sometimes the terms inside of the binomials that you're squaring will already have exponents attached to them. But fortunately, this doesn't change anything about the problem. We can still use our special product formulas to find these squares of binomials much faster than using the FOIL method. So, let's take a look at our problem. We have x squared plus one. All of that is going to be squared. So, in other words, we still have the form of a binomial that is squared. So, we have one term plus another at square and in this case, the a is our x squared and b is our one. So, all we have to do is just figure out our a and a squared and b and b squared, those four things, basically just put together our final answer. Alright? So, again, I like to sort of make a little table here. What's our a? Our a is just this entire first term over here, including the exponents. So, a is already x squared. So, what does that mean about a squared? We're just going to take x squared, which is our a term, and square that again. We have a power on top of the power, we multiply those exponents, which means that a squared is x to the fourth power. Alright? What about b and b squared? Our b is just going to be one, it's even more straightforward, and b squared is just going to be one squared, which again is just one. So, nothing really changes there. Alright. So, what happens here is now that we have these four things, we can go ahead and put together our final answer that's going to be in this format. Alright. So, what's our final answer? Well, we start with our A squared. A squared is what? Is it x squared or is it x to the fourth? It's x to the fourth. So, this is going to be x to the fourth plus or two times a times b. Take it slow. A in this case is x squared, so write that first in parentheses. Don't try to simplify this altogether because sometimes it confuses yourself, and then our b is just going to be one, and then plus our b squared, which is just going to be one squared, or you can just write this as one because we already figured out that this is just one. Okay. So, our final answer ends up being x to the fourth plus, now two times one doesn't change, it's just just two, and then two times x squared is going to be two x squared. So, that's what our middle term ends up becoming. Then finally, at the end, we just have a one over here. Alright. Now, if you don't trust me, you can go ahead and foil this out and you'll see that you get this exact same result. It's very similar as if you had something like x plus one squared. This is how you have to have an x squared at the beginning. Alright. So, this is our answer. Let me know if you have any questions, and thanks for watching.

Square each binomial. ( x 2 + 1 ) 2 (x^2+1)^2

x 4 + 2 x 2 + 1 x^4+2x^2+1 x 4 + 2 x 2 + 1

x 2 + 2 x 2 + 1 x^2+2x^2+1 x 2 + 2 x 2 + 1

x 2 + 2 x + 1 x^2+2x+1 x 2 + 2 x + 1

6. Exponents, Polynomials, and Polynomial Functions

Special Products

Intermediate Algebra

6. Exponents, Polynomials, and Polynomial Functions

Special Products

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Multiple Choice

Square each binomial.
$open paren x squared plus 1 close paren squared$

$x^4+2x^2+1$

$x^4+1$

$x^2+2x^2+1$

$x^2+2x+1$

Verified step by step guidance

Recognize that the expression $ (x^2 + 1)^2 $ means you are squaring the binomial $ x^2 + 1 $. This is equivalent to multiplying $ (x^2 + 1) $ by itself: $ (x^2 + 1)(x^2 + 1) $.

Apply the distributive property (also known as FOIL for binomials) to multiply each term in the first binomial by each term in the second binomial: $ x^2 \times x^2 $, $ x^2 \times 1 $, $ 1 \\times x^2 $, and $ 1 \\times 1 $.

Calculate each product separately: $ x^2 \\times x^2 = x^{2+2} = x^4 $, $ x^2 \\times 1 = x^2 $, $ 1 \\times x^2 = x^2 $, and $ 1 \\times 1 = 1 $.

Combine the like terms from the middle two products: $ x^2 + x^2 = 2x^2 $.

Write the final expanded expression by adding all the terms together: $ x^4 + 2x^2 + 1 $.

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