polynomials please help (1 Viewer)

[AFGHAN22]

A polynomial P(x) is odd, i.e. P(-x) = - P(x)
1. prove that P(0) = 0 and hence show that p(x) is divisible by x.
2. if b is a zero of P(x), show that –b is also a zero of P(x).
3. a polynomial P(x) is known to be odd, to be monic and have a zero -2. Show that p(x) is of at least degree 3.
State the paticukar polynomial Q(x) of degree 3 with the above properties. Is Q(x) unique?
4. A polynomial s(x) is odd, monic and has a zero -2. State the most general form of s(x) with degree 4</ d </ 6
note: </ denotes is less than or equal to

answers
3. Q(x) = 1x(x+2) (x-2) unique
4. s(x) = 1x (x-2)(x+2) (x-b)(x+b) where b is not equal to 2 or -2 , degree must be 5 since if b is zero of s(x) then so is –b

 

[insert-username]

A polynomial P(x) is odd, i.e. P(-x) = - P(x)
1. prove that P(0) = 0 and hence show that p(x) is divisible by x.

If P(-x) = - P(x), when x = 0, P(0) must equal - P(0) The only time when a number equals its negative is when the number is zero, i.e. 0 = -0.

For divisibility, when you substitute a value into P(x), then you will end up with 0. When x=0, P(x) = 0, therefore P(x) is divisible by x. I'm unsure about this proof, but it's all I can think of.


2. if b is a zero of P(x), show that –b is also a zero of P(x).

P(-x) = - P(x)

Therefore P(-b) = - P(b)

But P(b) = 0

Therefore P(-b) = 0

Therefore -b is a zero of the polynominal P(x) if P(x) is an odd function and P(b) is a zero.


3. a polynomial P(x) is known to be odd, to be monic and have a zero -2. Show that p(x) is of at least degree 3.

P(x) is an odd function, therefore P(0) = 0 (1 root)

P(-2) = 0 (2 roots)

Therefore P(2) = 0 (3 roots)

Therefore, since P(x) has at least 3 zeroes, it is of at least degree 3.


State the paticukar polynomial Q(x) of degree 3 with the above properties. Is Q(x) unique?

Q(x) = 0. Therefore one factor is x (when x is zero, the entire function is zero)

Q(2) = 0. Therefore another factor is (x-2) (same reason as above)

Q(-2) = 0. Therefore the third factorc is (x+2) (same reason)

Since Q(x) is of degree three, there are only 3 factors. It is monic, so there are no coefficients for x, and x must be positive (if it was negative, it would have a coefficient of -1 I think) so the polynomial is unique.

Q(x) = x(x-2)(x+2)


4. A polynomial s(x) is odd, monic and has a zero -2. State the most general form of s(x) with degree 4

S(x) cannot be of degree 4. Degree 4 functions are never odd functions since when you multiply a number by itself 4 times, it is always positive.


I_F

 

[AFGHAN22]

Thanks for your reply greatly appreciated thanks a lot man