Polynomial Question (1 Viewer)

[cutemouse]

Consider two polynomials P(x) and F(x)

When P(x) is divided by the remainder is 2x-11
When F(x) is divided by the remainder is x+4
With each division the quotient is the same.

(i) Show that P(x) and F(x) much have the same degree.

(ii) Find an expression for P(x).F(x)

(ii) Find the remainder when P(x) is divided by x+4.

 

[kaz1]

(i) Not sure but since divided by the same expression the remainder has the same degree it should be the same degree.
For (iii)
P(x)=(x+4)(x+2)Q(x) + 2x-11
P(-4)=-19

 

[cutemouse]

since divided by the same expression the remainder has the same degree it should be the same degree.

Hmm, if that was the answer then the question would ask something like "Explain why..." wouldn't it?

 

[kaz1]

Hmm, if that was the answer then the question would ask something like "Explain why..." wouldn't it?

Yeah true. If that's the case I have no idea.

 

[Prashant Sallan]

the quotient ALWAYS has to be one degree higher then the remainder.. and in this cause both remainders are linear, so both p(x) and f(x) should be a quadratic? (sorry, i know im explaining.. lol)

 

[study-freak]

 

[youngminii]

P(x) = (x^2 + 6x + 8)Q(x) + 2x - 11
F(x) = (x^2 + 6x + 8)Q(x) + 6x + 8
P(x).F(x) = [(x^2 + 6x + 8)Q(x) + (2x - 11)][(x^2 + 6x + 8)Q(x) + (6x + 8)]
= (x^2 + 6x + 8)^2.Q(x)^2 + (6x + 8)(x^2 + 6x + 8)Q(x) + (2x - 11)(x^2 + 6x + 8)Q(x) + (2x - 11)(6x + 8)
= (x^2 + 6x + 8)^2.Q(x)^2 + (4x - 3)(x^2 + 6x + 8)Q(x) + (12x^2 - 50x + 88)
?
That looks really dirty..

 

[Trebla]

Consider two polynomials P(x) and F(x)

When P(x) is divided by the remainder is 2x-11
When F(x) is divided by the remainder is x+4
With each division the quotient is the same.

(i) Show that P(x) and F(x) much have the same degree.

For (i), the key is that the quotients of both of P(x) and F(x) when divided by the same divisor are the same.

ANY polynomial divided by a quadratic MUST have a linear remainder in the form ax + b (note that the remainder can be constant if a = 0), so the remainder expression itself is somewhat irrelevant in this part.

So: (note the Q(x) expression is identical in both equations)
P(x) = (x² + 6x + 8)Q(x) + 2x - 11
F(x) = (x² + 6x + 8)Q(x) + x + 4

Let p be the degree of Q(x), so the degree of P(x) is (p + 2) and the degree of F(x) is (p + 2) as well, because multiplication of a polynomial of degree a with a polynomial of degree b, gives a polynomial of degree a + b.

To see this a bit more rigorously let the quotient be a general polynomial of degree p:
Q(x) = a_px^p + a_{p - 1}x^{p - 1} + .... + a₂x² + a₁x + a₀

We then get:
(x² + 6x + 8)(a_px^p + a_{p - 1}x^{p - 1} + .... + a₂x² + a₁x + a₀)
= (a_px^p+2 + a_{p - 1}x^{p + 1} + .... + a₂x⁴ + a₁x³ + a₀x²)
+ 6(a_px^p+1 + a_{p - 1}x^p + .... + a₂x³ + a₁x² + a₀x)
+ 8(a_px^p + a_{p - 1}x^{p - 1} + .... + a₂x² + a₁x + a₀)
= a_px^p+2 + (a_{p - 1}+6a_p)x^{p + 1} + ........ + (6a₀ + 8a₁)x + 8a₀

Therefore:
P(x) = a_px^p+2 + (a_{p - 1}+6a_p)x^{p + 1} + ........ + (6a₀ + 8a₁)x + 8a₀ + 2x - 11
= a_px^p+2 + (a_{p - 1}+6a_p)x^{p + 1} + ........ + (6a₀ + 8a₁ + 2)x + 8a₀ - 11
=> The degree of P(x) is p + 2

and

F(x) = a_px^p+2 + (a_{p - 1}+6a_p)x^{p + 1} + ........ + (6a₀ + 8a₁)x + 8a₀ + x + 4
= a_px^p+2 + (a_{p - 1}+6a_p)x^{p + 1} + ........ + (6a₀ + 8a₁ + 1)x + 8a₀ + 4
=> The degree of F(x) is p + 2

Therefore the degrees of P(x) and F(x) are equal.

This is probably as rigorous as I can get with this without restating the the very thing that is being proved...