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Find all integers k so that the trinomial can be factored using the methods of this section.$$3 y^{2}+k y+4$$...

Question

Find all integers k so that the trinomial can be factored using the methods of this section.$$3 y^{2}+k y+4$$

Find all integers k so that the trinomial can be factored using the methods of this section. $$ 3 y^{2}+k y+4 $$



Answers

Find all integers $k$ so that the trinomial is a perfect square trinomial. $$ k x^{2}+8 x y+y^{2} $$

We have 64 X squared minus six x plus k. So this is can be a spur. Less are We have my in the same here. Mhm and still a B plus this world when you divide the 16 by two So we have a B. It is equals toe eight x when you set a equals toe eight x Finished Worried We have 60 for expert so we must be equals toe one so okay, is equals to one.

We're being asked to factor the given expression. Well, you should always look for a greatest common factor first, and in this case, our greatest common factor will be K. Let's divide Eastern by K. Well, four k to the third. Divided by K is four case where negative four K squared, divided by K is negative four k and lastly, nine k divided by K is positive night. Now we're gonna look to see if we can factor fervor. Well, it looks like it might be a perfect square. Try Nobile, since both the first and last terms are perfect squares. So I was trying factor this such So I would bring down my gray this common factor and that is a perfect square trying o meal. It will faster into one binomial that gets square now, To get the first term in a binomial, you have to think what model meal when squared, would be four case where and it would be to K since UK times two k is four k square. And to get the last term in a binomial, you have to think of what number one square this night. And that's gonna be three and then we'll always take the sign of our middle term so minus. But now we have to confirm is in fact a perfect square triangle meal. To do that, we have to take twice the product of the two terms arrived by in our binomial and has to give us our middle term well. Two times two k times Negative three Well, two times two K is four K and four k times. Negative three is negative 12 k which is not our middle term. Therefore, this was not a perfect square China meal, so it can't factor into one binomial get squared. So I'm gonna erase everything here. So the only everything that we could try is to see if we can factor it by grouping. So what we would dio is multiply our first and last coefficients so we would do four times nine, which is equal to 36 and then we have to try and find the pair of factors that multiply the 36 that will give us our that will add to our middle coefficient of negative four. Well, let's think about our factors one and 36. But that won't work. Two in 18 that won't work three and 12. That won't work for nine. That won't work, and six and six that also won't work. Therefore, this China meal cannot be factored any fervor. So our final answer is K Times the Quantity a four K squared minus four k plus night.

So in this problem for being asked to find all the possible into your values of K that would allow us to factor this. Try no meal. Well, in order to do this, we will factor by grouping. So we would start by multiplying five times negative one while five times negative one is equal to negative five. So, typically, from here we look for are pairs of our factors of negative five that would somehow add to our middle coefficient K. So in this case, we have to put all of our possible combinations. Well, we could have positive one and negative five. We could have negative one and positive five. But there are no other values that multiplied to negative five. So now what we just need to do to find the values OK is we're going to add both of these sets up Well, one time Sorry one. The plus negative five is negative. Four and negative One plus five is positive for so are possible. Values for K are negative for and positive for

So in this problem, what we're trying to do is find all of the possible into your values for K that would allow us to factor the given. Try no meal. Well, to do this with a factor by grouping. So we would start by multiplying our first alas, coefficients to and negative three, which in this case is equal to negative six. Then we would try and figure out what factors multiply the negative six that would add to our middle coefficient of K. So we're trying to find all the possible values for K. We need to start by finding all the possible factors and negative six. Well, first, we would have won a negative six because I have negative one in positive six, which could have to a negative three. And we could have three. And negative too. Our sorry. Yep. Three negative to probably should have stuck with the same order as I did before. And these are all the values that could possibly multiply the negative six. So they get all of our possible K values. We just have to find the sum of each of these pairs Well, one plus negative six is negative five Negative one plus six is positive. Five Tu minus three is negative. One and three minus two is positive one. So now we found all the possible values for K Their negative five positive five negative one and positive one.


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