polynomials/misc qs (1 Viewer)

[totallybord]

hi,
could someone help me with these qs?
1) why is it true that the equation y=mx+c of a straight line is NOT a polynomial ?
2) is this answer rite?: the highest degree of the remainder of a long division if the divisor is a quadratic polynomial is 3 (the answer says one)
3) the polynomial (3a+b-1)x^3 + (a-2b)x has 4 zeroes. find the values of a n b.
the answers for a=2/7 and b=1/7 which makes the whole equation 0. i no that its impossible to have 4 zeroes bt how do i THINK like that so that i can solve it in the first place??
4) whats the difference btwn : y=ln(1-x/1+x) , y = ln(1-x^2) and y =ln (x^2-1) in terms of graph n everything else

i think thats it for today
anyway thanks in advance!

 

[A High Way Man]

maybe because its an equation, not an expression?

 

[vds700]

1) mx + c = 0 is a first degree polynomial

2) dont know what u mean

3) This question makes little sense. How can a cubic polynomial have 4 roots? A polynomial of degree n cannot have more than n roots.

4)
y = ln(1-x/1+x) = ln(1-x) - ln(1+x)
y = ln(1-x^2) = ln[(1+x)(1-x)] = ln(1+x) + ln(1-x)
y = ln(x^-1) = ln[(x+1)(x-!)] = ln(x+1) + ln(x-1) ....to graph these i guess u would use addition of ordinates, i dno, havent done graphs yet, or maybe u could differentiate them to find stationary points and points of inflexion

 

[foram]

maybe the polynomial has imaginary roots. (don't blame me if i don't know wat i'm talking about)

 

[vds700]

maybe the polynomial has imaginary roots. (don't blame me if i don't know wat i'm talking about)

A polynomimal of degree n has noi more than n complex roots (i think). Complex numbers include real and imaginary numbers as well, so this cunbic equation can't have 4 roots.

 

[tommykins]

1) why is it true that the equation y=mx+c of a straight line is NOT a polynomial ?

It is a polynomial of degree 1.

2) is this answer rite?: the highest degree of the remainder of a long division if the divisor is a quadratic polynomial is 3 (the answer says one)

The answer is one. If you divide x^3 by x^2 (degree 3 by degree 2) you will get degree 1.

3) the polynomial (3a+b-1)x^3 + (a-2b)x has 4 zeroes. find the values of a n b.
the answers for a=2/7 and b=1/7 which makes the whole equation 0. i no that its impossible to have 4 zeroes bt how do i THINK like that so that i can solve it in the first place??

Impossible (AFAIK), polynomial of n degree only have max n complex and real roots.

4) whats the difference btwn : y=ln(1-x/1+x) , y = ln(1-x^2) and y =ln (x^2-1) in terms of graph n everything else

Different domains etc. I suggest you download Graphmatica and sketch these out.

 

[Slidey]

hi,
could someone help me with these qs?
1) why is it true that the equation y=mx+c of a straight line is NOT a polynomial ?

It is a polynomial. It has degree 1. Anybody who says it isn't a polynomial is lying. It is the only linear polynomial (possibly excepting the trivial polynomial of degree 0 which is y=c).

4) whats the difference btwn : y=ln(1-x/1+x) , y = ln(1-x^2) and y =ln (x^2-1) in terms of graph n everything else

y=ln([1-x]/[1+x]) = ln(1-x) - ln(1+x)
y=ln(1-x^2) = ln([1-x][1+x]) = ln(1-x) + ln(1+x)
y=ln(x^2-1) = ln(x-1) + ln(1+x)

As you can tell they're all similar. There are subtle differences though - none of the above can be simplified into one of the others can they?

e.g. number one has asymptotes at x=-1 (bottom can't equal 0) and x=1 (ln0 doesn't exist so x must be less than 1) thus this graph will be between -1 < x < 1
Number 2 has asymptotes at the same points for the same reason. However, it looks different obviously - check its stationary points through the derivative to see why.
Number 3: x^2 is always positive. Thus the only values not possible are between -1 < x < 1. Thus function thus exists everywhere the other two don't. Heh.

Check them out on graphmatica.

 

[totallybord]

well according to terry lee, y=mx+c is not a polynomial. its true that equations cannot be polynomials but this one isnt exactly an equation rite?
q2 and q3 were in terry lees books too.
q2) tommykins, the divisor is the thing that u divide the eqn by such as 15/ 4 => 4 is the DIVISOR n 15 is the DIVIDEND
if u have any eqn and you divide by a quadratic, the hihgest remainder is of degree 3 coz otherwise if u have a degree of 4 or higher, u can still divide...isn't that rite?
q3) i thought it was impossible too...maybe because it's impossible so everything just equals 0

 

[tommykins]

well according to terry lee, y=mx+c is not a polynomial. its true that equations cannot be polynomials but this one isnt exactly an equation rite?
q2 and q3 were in terry lees books too.
q2) tommykins, the divisor is the thing that u divide the eqn by such as 15/ 4 => 4 is the DIVISOR n 15 is the DIVIDEND
if u have any eqn and you divide by a quadratic, the hihgest remainder is of degree 3 coz otherwise if u have a degree of 4 or higher, u can still divide...isn't that rite?
q3) i thought it was impossible too...maybe because it's impossible so everything just equals 0

I didn't really get the wording of q2), so I only explained what i could.

 

[totallybord]

thanks for replying anyway

 

[vds700]

the only way i can see 3 being possible is if a or b is a multiple of x. then if u expand, it will be a fourth degree equation with 4 roots

 

[Iruka]

I think the trick with Q3 is that since the polynomial, which appears to be a cubic, has four roots, this means that it must actually be the zero polynomial. So you have to find values for a and b which make both the coefficients zero.

 

[Slidey]

well according to terry lee, y=mx+c is not a polynomial. its true that equations cannot be polynomials but this one isnt exactly an equation rite?

y=mx+c is a polynomial. Ignore Terry Lee. Either he made a mistake or you are misreading the book. Either way, y=mx+c is a polynomial.

If y=mx+c was not a polynomial, the polynomials would not be a vector space, as it is not closed.

E.g. f(x)=x^2+x+1, g(x)=x^2
f(x)-g(x)=x+1 but by the definition of a vector space, it must be closed under addition/subtraction (h(x)=x+1 must also be a polynomial), thus either polynomials don't form a vector space, or the straight line is a polynomial.

It's basically a definition problem. For the sake of simplicity and consistency in maths, y=mx+c should definitely be classified as a polynomial.

I mean hell, every polynomial is a product of polynomials of the form y=mx+c anyway, right?

More here: http://en.wikipedia.org/wiki/Linear#Linear_polynomials

q3) i thought it was impossible too...maybe because it's impossible so everything just equals 0

From my limited studies of linear algebra, it sounds like Iruka is right - it is the 0 polynomial.