Polynomials question (1 Viewer)

[enigma_1]

p(x) is an odd polynomial of degree 3. It has x+4 as a factor, and when it is divided by x-3, the remainder is 21. Find p(x).

 

[Carrotsticks]

p(x) is an odd polynomial of degree 3. It has x+4 as a factor, and when it is divided by x-3, the remainder is 21. Find p(x).

Since P(x) is odd, let it be P(x) = ax^3 + bx

It has x+4 as a factor, so P(-4) = 0. (factor theorem)

It has remainder 21 when divided by x-3, so P(3) = 21 (remainder theorem)

You should have two expressions in terms of 'a' and 'b'.

Solve simultaneously.

 

[mr.habibbi]

Since P(x) is odd, let it be P(x) = ax^3 + bx

It has x+4 as a factor, so P(-4) = 0. (factor theorem)

It has remainder 21 when divided by x-3, so P(3) = 21 (remainder theorem)

You should have two expressions in terms of 'a' and 'b'.

Solve simultaneously.

where is this equation from ax^3 + bx

 

[Velocifire]

2013 guys

 

[Drdusk]

where is this equation from ax^3 + bx

Because it says P(x) is an odd polynomial. With an odd function P(x) = -P(x).

It also has degree 3. You can only have a degree 3 polynomial of the form ax^3 + bx that is an odd function because







If you add in any of the other terms such as x^2 or x or the constant, you won't have an odd function. For example if you take P(x) = ax^3 + bx^2 +cx then you get



It's how you do these questions. You write out a general equation of P(x) where the values of a, b, c... and so forth being random numbers.

 

[mr.habibbi]

Because it says P(x) is an odd polynomial. With an odd function P(x) = -P(x).

It also has degree 3. You can only have a degree 3 polynomial of the form ax^3 + bx that is an odd function because







If you add in any of the other terms such as x^2 or x or the constant, you won't have an odd function. For example if you take P(x) = ax^3 + bx^2 +cx then you get

ok i get this but the thing is where did you pull this equation from

what is the formula for odd polynomial or even polynomial for degree 2, 4, 5 so on...

 

[B1andB2]

ok i get this but the thing is where did you pull this equation from

what is the formula for odd polynomial or even polynomial for degree 2, 4, 5 so on...

for a function, if f(-x)=-f(x), then the function is odd. If f(-x)=f(x), then it is even

 

[B1andB2]

i remember this question from cambridge... tripped me up a lot lol

 

[ultra908]

u can prove that necessarily that the polynomial is ax^3+cx
Let f(x) = ax^3+bx^2+cx+d, a third degree polynomial
Since f(x) is odd, f(-x)=-f(x), i.e. -ax^3+bx^2-cx+d=-ax^3-bx^2-cx-d
Thus bx^2+d=0
Thus f(x)=ax^3+cx

 

[mr.habibbi]

u can prove that necessarily that the polynomial is ax^3+cx
Let f(x) = ax^3+bx^2+cx+d, a third degree polynomial
Since f(x) is odd, f(-x)=-f(x), i.e. -ax^3+bx^2-cx+d=-ax^3-bx^2-cx-d
Thus bx^2+d=0
Thus f(x)=ax^3+cx

This is exactly the part i don't get
how do you go from
bx^2+d = 0
to
f(x) = ax^3+cx

bx^2+d doesn't equal to ax^3+cx does it not?

 

[Drdusk]

This is exactly the part i don't get
how do you go from
bx^2+d = 0
to
f(x) = ax^3+cx

bx^2+d doesn't equal to ax^3+cx does it not?

The equation initially is f(x) = ax^3 + bx^2 +cx + d. Which rearranged is f(x) = ax^3 + cx + (bx^2 + d).
If you sub in bx^2 + d = 0 into that equation what remains is ax^3 + cx.

 

[mr.habibbi]

The equation initially is f(x) = ax^3 + bx^2 +cx + d. Which rearranged is f(x) = ax^3 + cx + (bx^2 + d).
If you sub in bx^2 + d = 0 into that equation what remains is ax^3 + cx.

thank you genius

 

[CM_Tutor]

Generalising, an even polynomial can only have terms with even powers and an odd polynomial can only have terms with odd powers. Further, if the degree of the polynomial is even then the polynomial cannot be odd but it can be even (though it could also be neither odd nor even). Similarly, if the degree of the polynomial is odd then the polynomial cannot be even but it can be odd (though it could also be neither odd nor even).

Taking :

• is even if and only if
• cannot be odd and a polynomial of degree

Taking :

• is odd if and only if
• cannot be even and a polynomial of degree

Any polynomial can be partitioned into two functions:

• , an even function, and
• , an odd function, such that
• . This can be achieved by defining:
• 
• 

In fact, this partition can be used on any function and if then the function is odd, and if then it is even. This partition is not useful for all functions. For example, is neither odd nor even and the partition does not yield a useful / sensible partition into an odd function and an even function. By contrast, is usefully separable into an odd component and an even component .

Applying this to the above problem, we need an odd function of degree 3. Starting from

we have

and we need the even part to be zero, which is how the above solutions established that . This can be shown as follows: