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Question

Factor each expression.
$$
\left(4 x^{2}+1\right)^{2}(2 x-1)^{-1 / 2}+16 x\left(4 x^{2}+1\right)(2 x-1)^{1 / 2}
$$

Amy Jiang · Accepted Answer

Do you see that we have a for where? Just one in common as well as a text, but it&#x27;s one terrible cool factor that up, so we&#x27;re left with a four x plus one finds a two x minus one Jabbar of negative 1/2 1 half so that&#x27;s minus one and invest 16 x.

Whitney Dillinger · Answer

in this problem, we&#x27;re going to factor the Tron Amiel Extra fourth plus X squared, minus two. Now I found Go ahead and look. I do not see any greatest common factors, so we can go ahead and weaken Factor are try no meal as usual. Since I do not have a coefficient here in front of my ex to the fore, I&#x27;m not going to be moving a number over two factor. We&#x27;re gonna factor as ISS. So let&#x27;s go ahead and let&#x27;s set up our two parentheses and let&#x27;s go ahead and start their eggs to the fourth. Well, there&#x27;s different ways I can get eggs to the fourth, but if we break it down into two parentheses evenly, we&#x27;re gonna get X squared times X squared. I noticed my middle term is positive and my last term is negative. It ordered for my last time to be negative. That means that one of these parentheses must be negative. So I&#x27;m gonna go ahead and make one parentheses. Positive. One parentheses. Negative. Now, in order to get this final number, I need to find two factors of negative, too, that when I add together, give me a sum of positive one. Well, there&#x27;s only two factors of negative, too. And those two factors are negative, too. And positive. One or another way you could do it is you could do positive. One. Nate. I&#x27;m sorry. Negative one Positive too. Well, negative. Two times one is negative. Two, but negative two plus one is negative one, so that will not work. Negative one. Tom&#x27;s positive two is negative to negative. One plus two is positive one. So my larger number has got to be positive said. In this parentheses I have to have plus two, and in the second parentheses, I have to have a negative one. So my factors are X squared, plus to an X squared minus one.

Matt Gibson · Answer

My common factor here would be two x squared and the quantity of X squared plus one square This one has square. This one is cubed so that would be my common factor. So when I divide that out, that leaves me with when this when I&#x27;m left with a two plus the quantity of X squared plus one, I would only have one cause I had three m subtracting my exponents so last I would just combine my leg terms here. So what have the two x squared times The X squared plus one squared and then the last would be to plus 23 plus X squared.

K B · Answer

So here we have four, actually fourth minus six x squared plus two and I&#x27;ll rewrite Were the factor this using grouping. So I&#x27;ll rewrite that minus six. Ex player as minus four. Extra weird bus. My eyes. Two arms form. So now we see that we want to do is factor each of them. So far, we have four x squared times expert minus one. And then here we have negative, too Times X squared minus one. So our solution to this question before X squared minus two. But before we even write the rest of it, we want to rewrite for expert minds, too, as two times X squared minus one and then the most expert minus one against what we actually have is too times X minus. Once word times X plus foreign sport. Here is our solution in this question.

Amrita Bhasin · Answer

we know we can pull 1/2 from both terms. 1/2 times X squared, minus one. We can further right? This is 1/2 times X squared, minus one squared. Remember the fear in the textbook that a squared plus B squared is a mite of speed times a plus B. The 1/2 still exists on the side, 1/2 times X minus, one times X plus one again, that&#x27;s that, Amos be tens a plus B.

Joshua Eastwood · Answer

that&#x27;s question as us to factor each expression. So we&#x27;re gonna start out by looking every time you factor for a common factor. That&#x27;s the first thing you want to look for. And to do that, you look at the numbers first and try to find the biggest number that will come out of that other words. What&#x27;s the biggest number that goes into four, eight and 16 and the answer is four. So we take that out and then we look at the exes and we notice there&#x27;s a three of them here X cube, so x x and X. There&#x27;s two here X squared, and there&#x27;s one here so we can take it. But one of them we can remove one axe, all right. And then what we do is we just divide each term. By that common factors. I&#x27;m gonna buy this buy for X. We divide this by four X, and I&#x27;m gonna divide this by four x and I look at that and I say four x Q divided by four. Actually using X squared eight divided by four is too expert about of axes X and then 16 divided by four is four an X divided by axis one. So I get here and then what we have to consider is does this factor. And in order to consider that we have to look at, I need two numbers that multiply to give me four. The FBI. Either one and four. We&#x27;re two and two that add up to this middle term, and in this case there aren&#x27;t any because one plus four is five. Two plus two is four, and if you&#x27;re negative, they have to both be paused for both the negative. For in your mouth you&#x27;re either positive or negative, but there&#x27;s no combinations of factors of four that will add to give you minus two. And what that means is that quadratic doesn&#x27;t factor, so that is my final and

Problem 56 Hard Difficulty

Factor each expression.
$$
\left(4 x^{2}+1\right)^{2}(2 x-1)^{-1 / 2}+16 x\left(4 x^{2}+1\right)(2 x-1)^{1 / 2}
$$

Answer

$\left(4 x^{2}+1\right)(2 x-1)^{1 / 2}\left[\left(4 x^{2}+1\right)(2 x-1)^{-1}+16 x\right]$

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$\left(4 x^{2}+1\right)(2 x-1)^{1 / 2}\left[\left(4 x^{2}+1\right)(2 x-1)^{-1}+16 x\right]$

Finite Mathematics and Calculus with Applications

Lily A.

Anna Marie V.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

Margaret L. Lial, Raymond N. Greenwell, Nathan P. RitcheyFinite Mathematics and Calculus with Applications

Lily A.

Anna Marie V.

Samuel H.

Joseph L.

Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Finite Mathematics and Calculus with Applications