Please, help me with this calculus: Consider the integral expression in x: P = x^3 + x^2 + ax + 1

[Felix_GG]

Q 2 Consider the integral expression in [imath]x[/imath]:

[imath]\qquad P = x^3 + x^2 + ax + 1[/imath]

...where [imath]a[/imath] is a rational number.

At [imath]a = \boxed{\; A \;}[/imath] the value of [imath]P[/imath] is a rational number for any [imath]x[/imath] which satisfies the equation [imath]x^2 + 2x - 2 = 0[/imath] and, in this case, the value of [imath]P[/imath] is [imath]\boxed{\; B \;}[/imath].

This is another problem i have never dealt with. I don't really know what i should start from.


Above is all my progress so far. And i'm afraid i don't quite undertand the problem, and what i should do.

 

[Dr.Peterson]

[imath]\qquad P = x^3 + x^2 + ax + 1[/imath]

...where [imath]a[/imath] is a rational number.

At [imath]a = \boxed{\; A \;}[/imath] the value of [imath]P[/imath] is a rational number for any [imath]x[/imath] which satisfies the equation [imath]x^2 + 2x - 2 = 0[/imath] and, in this case, the value of [imath]P[/imath] is [imath]\boxed{\; B \;}[/imath This is another problem i have never dealt with. I don't really know what i should start from. [ATTACH=full]35878[/ATTACH] Above is all my progress so far. And i'm afraid i don't quite undertand the problem, and what i should do. [/QUOTE] I don't think I've seen the term "integral expression" before, but [URL='https://en.wikipedia.org/wiki/Algebraic_fraction#Terminology']apparently[/URL] it just means it's written without any division. I don't see any need for calculus here. Just plug in the two solutions for the quadratic equation into the cubic P, and see what "a" has to be in order for the value to be rational. (I used synthetic division to evaluate it.)[/imath]

 

[NotJeffM]

There are two slightly different ways to think about this problem,

Perhaps the most obviuos way is to think that P names a family of cubic functions in x distinguished by a single rational parameter a. But we are immediately asked to consider only those members of the family where x satisfies a specific equation. That specific equation is a quadratic and, as you correctly found out, gives two solutions, namely [imath]x_1 = -1 + \sqrt{3}[/imath] and [imath]x_2 = - 1 - \sqrt{3}[/imath].

What is a little strange in this problem is that, having specified x, we still have not identified any specific cubic function. Instead, we have isolated within the family P of cubics two much more tightly defined families, which we can call U, associated with [imath]x_1[/imath], and V, associated with [imath]x_2[/imath]. It helps to name things.

With me so far?

Let’s concentrate first on the cubics of type U. Exactly how are they more tightly defined?

[math]U = (-1 + \sqrt{3})^3 + (-1 + \sqrt{3})^2 + a(- 1 + \sqrt{3}) + 1 \implies \\ U = (-1 + 3 * 1 * \sqrt{3} + 3 * (-1)(3) + 3\sqrt{3}) + (1 + 2 (-1) * \sqrt{3} + 3) - a + a\sqrt{3} + 1 \implies \\ U = -1 - 9 + 1 + 3 - a + 1 + (3 -2 - a)\sqrt{3} \implies\\ U = - 5 - a + (1 - a)\sqrt{3}. [/math]
Any difficulties so far?

Note that U is simply a linear function of a. We would get to the same place if we thought of P as function in two variables, x and a. Then it would be natural to find that once we specified x we are dealing with a function of one variable, namely a.

What is V?

 

[Steven G]

Maybe I missed something (I just woke up) but I don't see a question.

 

[blamocur]

There are two slightly different ways to think about this problem,

Perhaps the most obviuos way is to think that P names a family of cubic functions in x distinguished by a single rational parameter a. But we are immediately asked to consider only those members of the family where x satisfies a specific equation. That specific equation is a quadratic and, as you correctly found out, gives two solutions, namely [imath]x_1 = -1 + \sqrt{3}[/imath] and [imath]x_2 = - 1 - \sqrt{3}[/imath].

What is a little strange in this problem is that, having specified x, we still have not identified any specific cubic function. Instead, we have isolated within the family P of cubics two much more tightly defined families, which we can call U, associated with [imath]x_1[/imath], and V, associated with [imath]x_2[/imath]. It helps to name things.

With me so far?

Let’s concentrate first on the cubics of type U. Exactly how are they more tightly defined?

[math]U = (-1 + \sqrt{3})^3 + (-1 + \sqrt{3})^2 + a(- 1 + \sqrt{3}) + 1 \implies \\ U = (-1 + 3 * 1 * \sqrt{3} + 3 * (-1)(3) + 3\sqrt{3}) + (1 + 2 (-1) * \sqrt{3} + 3) - a + a\sqrt{3} + 1 \implies \\ U = -1 - 9 + 1 + 3 - a + 1 + (3 -2 - a)\sqrt{3} \implies\\ U = - 5 - a + (1 - a)\sqrt{3}. [/math]
Any difficulties so far?

Note that U is simply a linear function of a. We would get to the same place if we thought of P as function in two variables, x and a. Then it would be natural to find that once we specified x we are dealing with a function of one variable, namely a.

What is V?

My SymPy script produces a slightly different expression: [imath]U = -1 -5 \pm (a+4)\sqrt{3}[/imath], and my NumPy script confirms it numerically.

 

[Dr.Peterson]

My SymPy script produces a slightly different expression: [imath]U = -1 -5 \pm (a+4)\sqrt{3}[/imath], and my NumPy script confirms it numerically.

If you meant [imath]P(-1\pm\sqrt{3}) = -a -5 \pm (a+4)\sqrt{3}[/imath], my synthetic division agrees.

Maybe I missed something (I just woke up) but I don't see a question.

The implied question is to fill in A and B.

 

[blamocur]

If you meant [imath]P(-1\pm\sqrt{3}) = -a -5 \pm (a+4)\sqrt{3}[/imath], my synthetic division agrees.


The implied question is to fill in A and B.

Ooops! Of course it is [imath]-a-5 \pm (a+4)\sqrt{3}[/imath] -- thanks for the correction.