How Do I Solve Polynomial Problem #19?

[yungman]

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

I work to this point and get stuck:
https://www.physicsforums.com/attachments/270843

Can anyone help me?

Thanks

 

[mfb]

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?

 

[yungman]

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.

 

[Mark44]

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).

 

[SammyS]

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:
https://www.physicsforums.com/attachments/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?

 

[yungman]

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.

 

[yungman]

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

 

[yungman]

I have been looking at the question, the only values of t to get 2 binomials is this:

Is this right?

 

[mfb]

Right.

 

[SammyS]

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 $?x^2 +?x + 25$

 

[mfb]

@SammyS: Why would you want it to be a perfect square? No one required rational roots or any other special requirement.

 

[yungman]

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

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

 

[SammyS]

@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.

 

[SammyS]

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?

 

[yungman]

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:

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?

 

[SammyS]

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.

 

[yungman]

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

 

[SammyS]

Just to be clear, you can also have binomial products such as $(3x-1)(2x-25)$, which gives $t=-77$.

 

[mfb]

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.

 

[yungman]

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

 

[yungman]

Any comment?

 

[Mark44]

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 $6(x - a)^2$.

 

[yungman]

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.