polynomial question (1 Viewer)

[Joshmosh2]

P(x) is a monic polynomial of the fourth degree. When P(x) is divided by x+1 and x-2 the remainders are 5 and -4 respectively. Given that P(x) is an even function ie. one where P(x)=P(-x)
a. Express it in the form p0+p1x^1+p2x^2+p3x^3+p4x^4

Well what I instantly thought was equating both by subbing in the zeroes, but what equation would I substitute it in?
Maybe P(x)=A(x)Q(x)+R(x)?
well what I know is P(-1) = 5
and P(2) = -4
and then at one point, I would have to use the condition P(x)=P(-x)
Any help please? thanks.

 

[Drongoski]

P(x) = x^4 - 8x^2 + 12 ???

 

[Joshmosh2]

P(x) = x^4 - 8x^2 + 12 ???

YES! can you please explain to me how you got that?

 

[panda15]

YES! can you please explain to me how you got that?

Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.

 

[Joshmosh2]

Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.

Ahh, just simple intuition. Thanks!

 

[Joshmosh2]

One more thing,
when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
I know that R(1) = 2, which is the answer, but what could be a nice proof?

 

[faisalabdul16]

One more thing,
when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
I know that R(1) = 2, which is the answer, but what could be a nice proof?

Why would you need a nice proof when the answer is done in like 5 seconds

 

[panda15]

One more thing,
when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
I know that R(1) = 2, which is the answer, but what could be a nice proof?

P(x)=Q(x)(x^2-1)+(3x-1)
P(1)=0+3-1
=2

 

[19KANguy]

Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.

How do you make the two simultaneous equations?

 

[BenHowe]

Once you've used the fact that the polynomial is monic, so the coefficient of the x^4 is 1, and that it is an even function, so there is no odd power of x, you can write p(x)=x^4+bx^2+d. Now the polynomial has only two unknowns, b and d, and you have two pieces of information p(-1)=5 and p(2)=-4. When you use this you end up getting the two simulataneous equations b+d=4 and 4b+d=-20. Find the value of b & d and bob's your uncle

 

[19KANguy]

thanks