1972 Polynomials (1 Viewer)

[taggs-sasuke]

Hey. Can someone please help me with this question? I have the solutions but I don't understand them. My questions are in bold. Thank you!

A function f is called odd if f(-x) = -f(x) for all x.

(i) Prove that every odd function is zero at x=0

f(-x) = -f(x) for all x
therefore f(0) = -f(0)
2f(0) = 0 What is the link between this step and the previous step? (I'm lost...)
f(0) = 0

(ii) Prove that every odd polynomial P(x) is divisible by x.

f(0) = 0
i.e. the remainder when f(x) is divided by (x-0) is 0
i.e. x is a factor of f(x)

(iii) The polynomial P(x) is known to be monic, to be an odd function, and to have a root at x = -5. Show that P(x) has degree no less than 3.

P(-5) = 0
P(-5) = -P(5) odd function
therefore -P(5) = 0
So P(5) = 0
therefore (x-5) is also a factor
P(x) = x(x+5)(x-5) at least

(iv) Find a polynomial Q(x) of degree 3 with the properties given in part (iii). State, with reasons, whether there are any other polynomials of degree 3 with these properties.

deg 3 Q(x) = x(x-5)(x+5)
= x(x^2 - 25)
= x^3 - 25x

Q(-x) = (-x)^3 - 25(-x)
= -x^3 + 25x
= -Q(x)

No other polynomials as Q(x) is monic and of degree 3 Can someone explain this answer?

(v) State the form of the most general polynomial with the properties given in part (iii) and with degree d in the range 4 < or = d < or = 6

P(x) = x(x+5)(x-5)(x^2 + a)
for constant a

What is degree d?
How was the range incorporated into the answer?
How was (x^2 + a) arrived at?

Thank you!

 

[tommykins]

Hey. Can someone please help me with this question? I have the solutions but I don't understand them. My questions are in bold. Thank you!

A function f is called odd if f(-x) = -f(x) for all x.

(i) Prove that every odd function is zero at x=0

f(-x) = -f(x) for all x
therefore f(0) = -f(0)
2f(0) = 0 What is the link between this step and the previous step? (I'm lost...)
f(0) = 0

Simply rearrange,
f(0) + f(0) = 0
2f(0) = 0

 

[taggs-sasuke]

OH! Thanks.

 

[taggs-sasuke]

Can someone answer the other questions?

*feeling hopeful*

Thanks

 

[Iruka]

For the next bit, show that if P(x) is an odd polynomial (presumably an odd polynomial is a polynomial consisting of odd powers of x), then it is an odd function. Then use what you have proved in part (i) about odd functions to show that if P(x) is an odd polynomial, then P(0)=0, hence, x divides P(x).

 

[lolokay]

(iii)
No other polynomials as Q(x) is monic and of degree 3 Can someone explain this answer?

(v) State the form of the most general polynomial with the properties given in part (iii) and with degree d in the range 4 < or = d < or = 6

P(x) = x(x+5)(x-5)(x^2 + a)
for constant a

What is degree d?
How was the range incorporated into the answer?
How was (x^2 + a) arrived at?

Thank you!

iii) since we have the 3 factors of the 3rd degree polynomial, any other polynomial must have the same factors. so, it must be a constant multiple of x(x-5)(x+5), eg. 3x(x-5)(x+5). but as P(x) is monic, the constant out the front con only be one, .'. no other polynomials

iv) d is just the degree of the polynomial (like how the previous parts used d=3). as d can only be 4,5,6 and has to be odd, then d=5
if this polynomial is R(x), then R(x) = P(x)Q(x), where Q(x) is a quadratic
for R(x) to be monic and odd, the quadratic must be monic and even, as
R(-x) = P(-x)Q(-x)
-R(x) = -P(x)Q(-x)
R(x) = P(x)Q(-x)
so Q(x) = Q(-x)
so the quadratic:
ax² + bx + c =[equivalent to] a(-x)² + b(-x) + c
= ax² - bx + c
so b = -b, 2b = 0, b=0 (by equating coefficients)
and also, a=1
so Q(x) = x² + c
(same as saying x² + a)

 

[taggs-sasuke]

Oh, wow. Thanks a lot!