How to solve this polynomial problem

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  • #1
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I have problem solving this. The question is #19 in the first attachment. My work is in the second attachment
binomial.jpg


I work to this point and get stuck:
View attachment 270843

Can anyone help me?

Thanks
 

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  • #2
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How many roots does a parabola have if it can be written as product of two binomials? How many does it have if it cannot?
 
  • #3
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How many roots does a parabola have if it can be written as product of two binomials? How many does it have if it cannot?
I really don't know, this is a question from my grand daughter and this caught me cold. I am ashamed!!!

I set the equation to zero to get two roots.
 
  • #4
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In the Quadratic Formula, if the discriminant (the expression under the radical, ##b^2 - 4ac##) is positive, there will be two distinct roots. If the discriminant equals zero, there is one repeated root. If the discriminant is negative, there are no real roots (two complex roots).
 
  • #5
SammyS
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I have problem solving this. The question is #19 in the first attachment. My work is in the second attachment
View attachment 270842

I work to this point and get stuck:
View attachment 270843

Can anyone help me?

Thanks
This appears to be from a set of exercises for a lesson on factoring.

There are several methods for factoring a trinomial such as this. What method or methods have your granddaughter studied?
 
  • #6
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This appears to be from a set of exercises for a lesson on factoring.

There are several methods for factoring a trinomial such as this. What method or methods have your granddaughter studied?
I have to get hold of her. thanks

I made a mistake on the denominators in my first post, it should be 12 instead of 2. I tried to edit that, but it asked to put in the relevant formula. I can't find how to write formula here. I used to be able to write formula so I don't have to write on paper and scan it in. Seems like it changed since 10 years ago when I was active here.
 
  • #7
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Can anyone give me a hint how to look at it? I talked to her, she have not even learn the b^2-4ac stuffs. I don't know why she's even given this problem.

I am really not interested in this, I am drowning on my C++ programming already, just want to resolve this and help her on this one. I haven't even have a chance to work on my C++ today yet!!!

Thanks
 
  • #8
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I have been looking at the question, the only values of t to get 2 binomials is this:
Binomial2.jpg


Is this right?
 
  • #10
SammyS
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I have to get hold of her. thanks

I made a mistake on the denominators in my first post, it should be 12 instead of 2. I tried to edit that, but it asked to put in the relevant formula. I can't find how to write formula here. I used to be able to write formula so I don't have to write on paper and scan it in. Seems like it changed since 10 years ago when I was active here.
I suppose that you are referring to the quadratic formula:
The solution to the general quadratic equation: ##ax^2+bx+c=0 ##
is given by ##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

So, yes, you left the factor of ##a## out of the denominator. In this case ##a=6##.

In addition to what @Mark44 said, if ##b^2-4ac## is a perfect square, then ##ax^2+bx+c ## can be written as the product of two binomials. For your (granddaughter's) problem, you need ##t^2-600## to be a perfect square, (as well as being positive, of course).
This problem may be done this way, but it will be quite cumbersome. There are at least 6 possible values for t.

Another way to do this is to systematically find all of the pairs of binomials product give ##6x^2## for the leading term and ##25## for the constant term.

For example: ##(2x + ?)\times(3x + ?)## gives ##6x^2 + ... ##

and ##(?x + 1)\times(?x + 25)## gives ## ?x^2 +?x + 25##
 
  • #11
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@SammyS: Why would you want it to be a perfect square? No one required rational roots or any other special requirement.
 
  • #12
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I suppose that you are referring to the quadratic formula:
The solution to the general quadratic equation: ##ax^2+bx+c=0 ##
is given by ##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

So, yes, you left the factor of ##a## out of the denominator. In this case ##a=6##.

In addition to what @Mark44 said, if ##b^2-4ac## is a perfect square, then ##ax^2+bx+c ## can be written as the product of two binomials. For your (granddaughter's) problem, you need ##t^2-600## to be a perfect square, (as well as being positive, of course).
This problem may be done this way, but it will be quite cumbersome. There are at least 6 possible values for t.

Another way to do this is to systematically find all of the pairs of binomials product give ##6x^2## for the leading term and ##25## for the constant term.

For example: ##(2x + ?)\times(3x + ?)## gives ##6x^2 + ... ##

and ##(?x + 1)\times(?x + 25)## gives ## ?x^2 +?x + 25##
I use your logic to have this, but I have 4 variables A, B, C and D, and I cannot solve it.
##(Ax+B)(Cx+D)=6x^2 + tx + 25##
It would be nice to do this instead of using

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

She have not learn this formula, it would be nice to have a way to do this without

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##
 
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  • #13
SammyS
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@SammyS: Why would you want it to be a perfect square? No one required rational roots or any other special requirement.
Looking at the other exercises in the image in the OP, I conclude that these exercises have to do with factoring trinomials into the product of binomials, all with integer coefficients.
 
  • #14
SammyS
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I use your logic to have this, but I have 4 variables A, B, C and D, and I cannot solve it.
##(Ax+B)(Cx+D)=6x^2 + tx + 25##
It would be nice to do this instead of using

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##
Correct. You have ##(Ax+B)(Cx+D)=6x^2 + tx + 25##.

But A and C are related: ##A\cdot C = 6##. Also ##B\cdot D = 25##.

Now get systematic.
With A = 1 and C = 6, let B = 1 and D =25, and multiply ##(x+1)(6x+25) ## to get ##t##.

Then let B = 5 and D =5, and multiply ##(x+5)(6x+5) ## to get another value for ##t##.

Then let B = 25 and D = 1 ... . Don't forget, B and D can both be negative.

Now consider A = 2 and C = 3. Run through the cases for B and D again.

When can you stop?
 
  • #15
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Correct. You have ##(Ax+B)(Cx+D)=6x^2 + tx + 25##.

But A and C are related: ##A\cdot C = 6##. Also ##B\cdot D = 25##.

Now get systematic.
With A = 1 and C = 6, let B = 1 and D =25, and multiply ##(x+1)(6x+25) ## to get ##t##.

Then let B = 5 and D =5, and multiply ##(x+5)(6x+5) ## to get another value for ##t##.

Then let B = 25 and D = 1 ... . Don't forget, B and D can both be negative.

Now consider A = 2 and C = 3. Run through the cases for B and D again.

When can you stop?
Ok, I expanded these out:
Binomial 3.jpg


I see the relation, I connected the pairs with same value of t. These are all the combination. But what does these means?

Does this means t = 25, 31, 35, 53, 77 and 151 all work?
 
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  • #16
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Ok, I expanded these out:
View attachment 270852

I see the relation, I connected the pairs with same value of t. These are all the combination. But what does these means?

Does this means t = 25, 31, 35, 77 and 151 all work?
Yes. Also 53 .

Additionally, t can be the negative of each of these.
 
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  • #17
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Yes. Also 53 .

Additionally, t can be the negative of each of these.
Thanks, I missed the 53 when I read my notes. It's there also. I put it back in the last post already.

Thanks
 
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  • #18
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Just to be clear, you can also have binomial products such as ##(3x-1)(2x-25)##, which gives ##t=-77##.
 
  • #19
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Looking at the other exercises in the image in the OP, I conclude that these exercises have to do with factoring trinomials into the product of binomials, all with integer coefficients.
I don't share that interpretation.
 
  • #20
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Thanks for the help.

Actually after sitting on it for a day, I really feel the answer in my post #8 is more accurate. I understand the answer in post #15 looks better with integer coef for x.

BUT, this is NOT the full set of answer. Remember, coef for both x and 25 can be fraction, the requirement is just to have the final coef of x to be 6 and final number is 25. Also, coef of x in the two binomial can be of different polarity also, that complicate the situation even more. The few values in post #15 is ONLY a very small SUBSET of the value t can take on.

I still believe the answer in post #8 is the ONLY correct answer. Grand daughter just has to learn the Quadratic Formula quick and use it.

Thanks
 
  • #21
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Any comment?
 
  • #22
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Any comment?
Since there isn't any restriction given in the problem statement about solutions being integers or rational numbers, the solution in post #8 looks fine to me.

However, I would simplify things a bit.
$$t \ge 10\sqrt 6 \text{ or } t \le -10\sqrt 6$$
If ##t = \pm 10\sqrt 6## then the binomial will have two repeated factors; i.e., it will have the form ##6(x - a)^2##.
 
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  • #23
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Since there isn't any restriction given in the problem statement about solutions being integers or rational numbers, the solution in post #8 looks fine to me.

However, I would simplify things a bit.
$$t \ge 10\sqrt 6 \text{ or } t \le -10\sqrt 6$$
If ##t = \pm 10\sqrt 6## then the binomial will have two repeated factors; i.e., it will have the form ##6(x - a)^2##.
Thanks, I'll tell her to read this thread.
 

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