Advanced Functions: ax^3+bx+c (a≠0 b≠0) has factor x^2+px+1. Show a^2-c^2=a

[Hikmat]

Advanced Functions: ax^3+bx+c (a≠0 b≠0) has factor x^2+px+1. Show a^2-c^2=a

ax^3 + bx + c (a≠0 b≠0) has a factor x^2 + px + 1. Show that a^2 - c^2 = ab.

I'm having trouble dealing with 4 unknowns, am I going about this the wrong way?

 

[mmm4444bot]

ax^3 + bx + c (a≠0, b≠0) has a factor x^2 + px + 1

Show that a^2 - c^2 = ab

I'm having trouble dealing with 4 unknowns, am I going about this the wrong way?

You haven't described any strategy or shown any work thus far, so I cannot comment on your "way".

Can you explain what you're thinking about, when you say, "trouble dealing with 4 unknowns". What is it that you're trying to do?

We don't need to know actual values for any of the symbols (the variable or the three parameters), if that's your concern. One way to begin involves polynomial long division. Have you learned how to do that?

In other words, can you divide [ax^3+bx+c] by [x^2+px+1], in order to get an expression for the remainder?

It would be helpful to get an expression for the remainder because we are told that [x^2+px+1] is a factor of [ax^3+bx+c].

Do you know that this means the remainder must equal zero?

Please answer the questions in this reply, and we can go from there.

If you already know how to perform polynomial long division, please show what expressions you get for the quotient and the remainder. :cool:

 

[Hikmat]

You haven't described any strategy or shown any work thus far, so I cannot comment on your "way".

Can you explain what you're thinking about, when you say, "trouble dealing with 4 unknowns". What is it that you're trying to do?

We don't need to know actual values for any of the symbols (the variable or the three parameters), if that's your concern. One way to begin involves polynomial long division. Have you learned how to do that?

In other words, can you divide [ax^3+bx+c] by [x^2+px+1], in order to get an expression for the remainder?

It would be helpful to get an expression for the remainder because we are told that [x^2+px+1] is a factor of [ax^3+bx+c].

Do you know that this means the remainder must equal zero?

Please answer the questions in this reply, and we can go from there.

If you already know how to perform polynomial long division, please show what expressions you get for the quotient and the remainder. :cool:

I have learned long division, but just by looking at it I can also tell that the quotient must be ax+c, because (ax+c)(x^2+px+1) will give me the same leading coefficient and constant as the original function ax^3 +bx+c.

I also know that by multiplying ax+c and x^2+px+1 that apx^2 + cx^2 is 0 (original has 0x^2). I can also see that ax+cpx=bx.

I currently have 3 equations.
apx^2 + cx^ = 0
ax+cpx=bx
and from the question: a^2 - c^2 = ab.

I think this may be the wrong method because using 4 equations for 4 unknowns would be if I was trying to get actual values. I don't know how to prove what's being asking in the question however.

 

[stapel]

I have learned long division, but just by looking at it I can also tell that the quotient must be ax+c, because (ax+c)(x^2+px+1) will give me the same leading coefficient and constant as the original function ax^3 +bx+c.

I also know that by multiplying ax+c and x^2+px+1 that apx^2 + cx^2 is 0 (original has 0x^2). I can also see that ax+cpx=bx.

I currently have 3 equations.
apx^2 + cx^ = 0
ax+cpx=bx
and from the question: a^2 - c^2 = ab.

I think this may be the wrong method because using 4 equations for 4 unknowns would be if I was trying to get actual values. I don't know how to prove what's being asking in the question however.

Try using long division, like the previous helper suggested: do the long division, and use the fact that the remainder must (by definition) equal zero. You should, fairly quickly, discover that two of the constants (a, b, c, and p) is zero, and that the other two are equal to each other. This simplifies things considerably.

 

[Hikmat]

Try using long division, like the previous helper suggested: do the long division, and use the fact that the remainder must (by definition) equal zero. You should, fairly quickly, discover that two of the constants (a, b, c, and p) is zero, and that the other two are equal to each other. This simplifies things considerably.

Thanks for the help everyone. It is infact much simpler than I thought, here is my solution.

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Since $\displaystyle x^2\, +\, px\, +\, 1$ is a factor of $\displaystyle ax ^3\, +\, bx\, +\, c$ , and since the other factor must be linear, in the form $\displaystyle ax\, +\, c$, then we have:

. . . . .$\displaystyle ax^3\, +\, bx\, +\, c\, =\, ax^3\, +\, (ap\, +\, c)\, x^2\, +\, (cp\, +\, a)\, x\, +\, c$

Subtracting the first and last terms from either side, we get:

. . . . .$\displaystyle bx\, +\, c\, =\, (ap\, +\, c)\,x^2\, +\, (cp\, +\, a)\, x$

The first equation says that $\displaystyle ap\, +\, c\, =\, 0$ and $\displaystyle cp\, +\, a\, =\, b.$ This says that $\displaystyle c\, =\, -ap$ so $\displaystyle -\frac{c}{a}\, =\, p.$

Plugging this into the second equation, we get:

. . . . .$\displaystyle c\, \left(-\frac{c}{a}\right)\, +\, a\, =\, b$

. . . . .$\displaystyle -\frac{c^2}{a}\, +\, a\, =\, b$

. . . . .$\displaystyle -\frac{c^2}{a}\, +\, \frac{a^2}{a}\, =\, b$

. . . . .$\displaystyle \frac{a^2\, -\, c^2}{a}\, =\, b$

Multiplying through, we get:

. . . . .$\displaystyle a^2\, -\, c^2\, =\, ab$

 

[stapel]

Thanks for the help everyone. It is in fact much simpler than I thought, here is my solution.

---

Since $\displaystyle x^2\, +\, px\, +\, 1$ is a factor of $\displaystyle ax ^3\, +\, bx\, +\, c$ , and since the other factor must be linear, in the form $\displaystyle ax\, +\, c$, then we have:

. . . . .$\displaystyle ax^3\, +\, bx\, +\, c\, =\, ax^3\, +\, (ap\, +\, c)\, x^2\, +\, (cp\, +\, a)\, x\, +\, c$

Subtracting the first and last terms from either side, we get:

. . . . .$\displaystyle bx\, +\, c\, =\, (ap\, +\, c)\,x^2\, +\, (cp\, +\, a)\, x$

The first equation says that $\displaystyle ap\, +\, c\, =\, 0$ and $\displaystyle cp\, +\, a\, =\, b.$ This says that $\displaystyle c\, =\, -ap$ so $\displaystyle -\frac{c}{a}\, =\, p.$

Plugging this into the second equation, we get:

. . . . .$\displaystyle c\, \left(-\frac{c}{a}\right)\, +\, a\, =\, b$

. . . . .$\displaystyle -\frac{c^2}{a}\, +\, a\, =\, b$

. . . . .$\displaystyle -\frac{c^2}{a}\, +\, \frac{a^2}{a}\, =\, b$

. . . . .$\displaystyle \frac{a^2\, -\, c^2}{a}\, =\, b$

Multiplying through, we get:

. . . . .$\displaystyle a^2\, -\, c^2\, =\, ab$

Or, using long division, we get a remainder of:

. . . . .$\displaystyle (b\, -\, a\, +\, ap)\, x\, +\, (ap)\, 1\, =\, 0$

The only way this equality is true for all values of $\displaystyle x$ is if each of the parentheticals on the left-hand side is equal to zero. The second parenthetical says:

. . . . .$\displaystyle ap\, =\, 0$

We are given that $\displaystyle a\, \neq\, 0,$, so it must be that $\displaystyle p\, =\, 0.$ Plugging this into the other parenthetical, we get:

. . . . .$\displaystyle b\, -\, a\, +\, 0\, =\, 0$

. . . . .$\displaystyle b\, =\, a$

When we did the long polynomial division, the factor across the top of the division was:

. . . . .$\displaystyle ax\, -\, ap\, =\, a(x\, -\, p)$

But since $\displaystyle p\, =\, 0,$ we really had a complete factorization in the form of:

. . . . .$\displaystyle ax\, (x^2\, +\, 1)\, =\, ax^3\, +\, ax\, =\, ax^2\, +\, bx\, +\, c$

This tells us that $\displaystyle c\, =\, 0.$ Since $\displaystyle a\, =\, b,$ then $\displaystyle a\, -\, b\, =\, 0\, =\, c$. Squaring both sides of this gives us:

. . . . .$\displaystyle a^2\, -\, 2ab\, +\, b^2\, =\, c^2$

. . . . .$\displaystyle a^2\, -\, c^2\, =\, 2ab\, -\, b^2$

But $\displaystyle a\, =\, b,$ so $\displaystyle b^2\, =\, ab$ and:

. . . . .$\displaystyle a^2\, -\, c^2\, =\, 2ab\, -\, ab\, =\, ab$