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Numerade Educator



Problem 23 Easy Difficulty

Perform the indicated operations.


$\therefore(x+2)^{2}=x^{2}+4 x+4$


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Video Transcript

and this problem we're being asked to multiply to try no wheels. To do this, we're going to use to the strip of the property. So here's how it works. We're gonna take the first term in our first try no meal, which is X squared, and we're gonna multiply it by our second try. No meal X squared plus X minus two. Then we're going to take the second term in our first try. No meal, which is minus X and also multiply it by that second China meal X squared plus X minus two. Then we're gonna take the last term in the first train. Oh, meal, which is positive, too. And also multiply that by our second China meal X squared plus X minus two. Now we just have to go ahead and distribute well X squared times X Where is extra before X squared Times X is positive X to the third an X squared times Negative too is negative two X square. Now we're gonna multiply negative x times X squared which is negative x to the third. Then we multiply Negative x times X, which is negative. X word. No, we multiply negative X times negative, too, which is positive to X. And lastly, we need to multiply two times X word, which is positive to X word we multiply two times X, which is positive to x and then we multiplied two times negative too, which is negative for now. We just have to simplify by combining our like terms. So let's start with our first term, which is excellent. For As you can see, it doesn't have a life term, so we simply bring it down. Now we're gonna combine their X to the third terms while we have positive X to the third and minus X to the third. Well, X to the third minus X to the third is zero. So those terms cancel out. Now we're gonna combine our X squared terms while we have negative two x squared minus X squared plus two x squared. Well, negative two minus one. Because, remember, there's a one for coefficient, so negative two minus one is negative. Three and negative three plus two is negative one. So we'll just have negative X squared. Remember, we don't have to put the one for a coefficient. Now let's combine our ex terms while we have two x plus two x, which is positive for X. And then lastly, we have our constant term minus four. Well, it doesn't have a late term, so we just bring it down. So our final answer