[SOLVED] Help Needed...

**Szpieg** · June 6th 2009, 04:53 AM

-Show that x-a is a factor of x^n-a^n for any positive integer n.

-Show that x+a is a factor of x^n+a^n for any positive odd integer n.

Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.

Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?

**yeongil** · June 6th 2009, 05:19 AM

Originally Posted by Szpieg

-Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.

The slope of the line y = x is 1. The slope of the line with the points (a, b) and (b, a) is
$\frac{y_2 - y_1}{x_2 - x_1} = \frac{a - b}{b - a} = -1$ ,
so this line is perpendicular to the line y = x. (Slopes of perpendicular lines are negative reciprocals of each other.)

Pick a point (c, c) that is on the line y = x. Use the distance formula to find the distance from (a, b) to (c, c), and from (b, a) to (c, c).

01

**Szpieg** · June 6th 2009, 05:40 AM

thanks for the help, if anyone can help we with the others, it would be much appreciated...

**yeongil** · June 6th 2009, 05:40 AM

Originally Posted by Szpieg

-Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?

If the constants a, b, c, d, e, f form an arithmetic sequence, then
b = a + n
c = a + 2n
d = a + 3n
e = a + 4n
f = a + 5n
for some n. Substitute:

$\begin{aligned} ax\ &+ &(a + n)y\ &=\ a + 2n \\ (a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n \end{aligned}$

Solve by elimination. Multiply the 1st equation by -(a + 3n)/a:

$\begin{aligned} -(a + 3n)x\ &- &\frac{(a + n)(a + 3n)}{a}y\ &=\ -\frac{(a + 2n)(a + 3n)}{a} \\ (a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n \end{aligned}$

Add the 2 equations to cancel out the x's. The y coefficients, added together, give us

$- \frac{(a + n)(a + 3n)}{a} + (a + 4n) = - \frac{a^2 + 4an + 3n^2}{a} + \frac{a^2 + 4an}{a} = -\frac{3n^2}{a}$ ,

and the constant terms, added together, give us

$- \frac{(a + 2n)(a + 3n)}{a} + (a + 5n) = - \frac{a^2 + 5an + 6n^2}{a} + \frac{a^2 + 5an}{a} = -\frac{6n^2}{a}$ .

The two equations added together, therefore, is

$-\frac{3n^2}{a}y = -\frac{6n^2}{a}$ ,

meaning that y = 2.

Substitute y = 2 into either of the original equations (I chose the first):

$\begin{aligned} ax + (a + n)y &= a + 2n \\ ax + (a + n)(2) &= a + 2n \\ ax + 2a + 2n &= a + 2n \\ ax &= -a \end{aligned}$

So x = -1. We therefore have one solution to the system of equations, (-1, 2).

01

**Szpieg** · June 6th 2009, 05:57 AM

OMG, thank you so much, this one i had no idea how to do=] ull be getting thanks from me =] and respect...by the way how old are you and what school do you go to?

PS, i got the same answer, but not in a true proof...i used multiple examples, but my teacher does not want that...this is perfect

**Prove It** · June 6th 2009, 06:23 AM

Originally Posted by Szpieg

-Show that x-a is a factor of x^n-a^n for any positive integer n.

-Show that x+a is a factor of x^n+a^n for any positive odd integer n.

-Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.

-Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

-Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?

For your first question, call your function $f(x)$ .

So $f(x) = x^n - a^n$ .

By the Factor and Remainder Theorems, $(x - a)$ is a factor IF $f(a) = 0$ .

So $f(a) = a^n - a^n = 0$ .

So $(x - a)$ is a factor of $x^n - a^n$ .

An alternative proof is done using factorisation.

For any $a, b$

$a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})$ .

Applying this knowledge to $f(x)$ we find

$f(x) = (x - a)(x^{n - 1} + x^{n - 2}a + x^{n -3}a^2 + \dots + x^2a^{n - 3} + xa^{n - 2} + a^{n - 1})$

So it can clearly be seen that $(x - a)$ is a factor.

For your second problem, to show $(x + a)$ is a factor of $x^n + a^n$ for any $n$ that is an odd positive integer, do the same thing.

Let $f(x) = x^n + a^n$ .

By the Factor and Remainder Theorems,
$(x + a)$ is a factor if $f(-a) = 0$ .

So $f(-a) = (-a)^n + a^n$ .

If $n$ is even, you have $f(-a) = a^n + a^n = 2a^n$ , so $x - a$ is not a factor if $n$ is even,

BUT If $n$ is ODD, you have $f(-a) = -a^n + a^n = 0$ .

So $(x + a)$ is a factor of $f(x)$ as long as n is a positive odd integer.

Alternatively, use a similar alternate proof to the one above.

If $n$ is an odd integer, then

$a^n + b^n = (a + b)(a^{n - 1} - a^{n - 2}b + a^{n - 3}b^2 - \dots + a^2b^{n - 3} - ab^{n - 2} + b^{n - 1})$ .

So $f(x) = x^n + a^n = (x + a)(x^{n - 1} - x^{n - 2}a + x^{n - 3}a^2 - \dots + x^2a^{n - 3} - xa^{n - 2} + a^{n-1})$ .

Here you can clearly see that $x + a$ is a factor of $f(x)$ , but only if $n$ is a positive odd integer.[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]

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