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4) A race Car tra velj 0 $ Conj +ant Jpee 4 around circu lar #rac k Uka 0o4 | d Aappe n to +he Cenfripe tal accele ration 0 f +he car 8 F #Rz spee d dou blesA) Decr...

Question

4) A race Car tra velj 0 $ Conj +ant Jpee 4 around circu lar #rac k Uka 0o4 | d Aappe n to +he Cenfripe tal accele ration 0 f +he car 8 F #Rz spee d dou blesA) Decrea sej by B) In crea Jes Byfac _ tor of |/) , factor of 4 .C)IncreaJef by factor of 2. 0) I+ stays +e Same,

4) A race Car tra velj 0 $ Conj +ant Jpee 4 around circu lar #rac k Uka 0o4 | d Aappe n to +he Cenfripe tal accele ration 0 f +he car 8 F #Rz spee d dou bles A) Decrea sej by B) In crea Jes By fac _ tor of |/) , factor of 4 . C)IncreaJef by factor of 2. 0) I+ stays +e Same,



Answers

Factor each polynomial completely. $$4 a^{2} b^{2}-4 a b w+w^{2}$$

Where we asked the factor that given Paulie Mobile. Well, anytime you're asked the factor, always look for a greatest common factor first, which in this case, there isn't one. Now I see that there's four terms, so we're gonna try and factor by group. So let's look at the 1st 2 terms and find the greatest common factor. Well, the greatest common factor of those 1st 2 terms is W Now, when I divide both terms to my W, we're gonna be left with a minus four. Now, we're gonna look at the 2nd 2 terms and find the greatest common factor. Well, the greatest common factors be, but because this term is negative, I'm gonna actually factor all that negativity. Now, when I divide both terms by negative be we're gonna be left with a minus four. And now this is great, because now we have this common factor of a minus four. So we're gonna factor out the A mine's four term, and then we're left with W minus B, which will be our next factor. Never factor could get after that

We have four ace. Where eastward See, like the six A East where it's C minus four a. C plus a look. Two greatest common factor. First, all terms have at least a two that could be divided from them. They all have at least one A. They don't have bees that a little have seeds. So to a is the greatest common factor, which by that from each term left with to a B squared C minus three beast where it see minus two C plus four.

We're being asked. The faster the given. Polly Nobile, Remember, any time you need to factor, you should always look for a greatest common factor. First, however, there wasn't one in this case. Now, as you can see, we have a try. No meal. Who's a term is not equal to one. So what we're going to need to do is we're gonna have to multiply are a term which is for by the coefficient of our last term, which is one which is going to give us four. So we have to figure out what's gonna multiply the four that will give us a coefficient of our middle term, which is negative for Well, I'm going to use the factors to and to, because if I make both of those factors negative, negative two times negative too will be positive for and negative to minus two is negative for, so this is perfect. So what I'm going to do is I'm going to replace this negative for a B turn with negative to a B and negative to a B, because that's equal to negative for E. B. So we're gonna have for B squared minus two AP minus two AP plus a square, and now we're gonna faster by grouping. So let's look at our 1st 2 terms and find the greatest common factor. Well, that's gonna be to be so when I divide both terms by to be, I'm gonna be left with to be minus a. Now, let's look at the 2nd 2 terms and find the greatest common factor. Well, they both have an A in common. And because the first term starts with a negative, I'm gonna actually factor out. Negative, eh? So when we divide both turns by negative A, we're gonna be left with to be minus a Well, this is perfect, because now they have the same common factor. So I can factor out the to be minus a And as you can see, we're also left with a factor off to be minus a so we can combine these and simply write our final answer as the quantity of to be minus a square

All right. This is a perfect square trying Amiel. It fits this pattern X squared Plus to exploit plus y squared so we can factor It is in this pattern. So what we'll do is take the squared of M minus end square, which is m minus. And and then we'll take the square to this four, which is to and when you do two times m minus and double, do you have four times M minus end. So that's why the factory ization is M minus and plus two all squared.


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