polynomial question =s (1 Viewer)

[akukei]

2004 HSC Question 3

(b) Let P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is −11.
When P(x) is divided by (x − 3) the remainder is 1.

(i) What is the value of b?
(ii) What is the remainder when P(x) is divided by (x + 1)(x − 3)?

I can't do part two. =/ Any help greatly appreciated.

 

[tommykins]

Hint: remainder when P(x) is divided by (x + 1)(x − 3) takes the form of Ax+B.

Answer should be 3x-8, post if you want the full solution.

 

[akukei]

so its just a(x + 1) + b? and you sub in a and b to get the answer?

Is that something we are just supposed to know?

 

[alez]

i dont get it
can u post full soln please...

 

[tommykins]

I believe it requires 4unit thinking, but 3unit knowledge is possible.

(b) Let P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is −11.
When P(x) is divided by (x − 3) the remainder is 1.

(ii) What is the remainder when P(x) is divided by (x + 1)(x − 3)?

Say we have a polynomial P(x). When we divide this polynomial by a non-factor polynomial, say g(x) where the degree of g(x) is 2, we get
P(x) = g(x).Q(x) + R(x)

Now, since the degree of g(x) is 2, division gives a remainder of degree 1.
Thus, R(x) takes the form of Ax+B where A and B are integers.

Subbing in x = -1, we know the remainder of P(x) in the above question is -11.
Subbing in x =3 , we know the remainder is 1

So that means
-A+B = -11 (note that you sub in x = -1 for the Ax+B)
and 3A+B = 1
Simultaneously solve you get A = 3, B = -8
Thus remainder is 3x-8.

 

[akukei]

So as a general rule, when you divide a polynomial P(x) by a non-factor polynomial g(x) of degree n, the degree of R(x) is always n-1? If that makes sense...

 

[tommykins]

if p(x) is of degree r and g(x) is of degree n where r > n, then im pretty sure division of P(x) by g(x) gives a remainder of degree n-1.

EDIT : don't think its ALWAYS, but mostly, depends on the polynomial.


yep. 99% sure in the 3u cases though.

 

[Rampager]

I did it this way:

P(3) = 4a - 11 = 1
a = 3
P(-1) = b = -11
b = -11

All polynomials take the form of P(x) = D(x)Q(x) + R(x)
where D(x) is the divisor, Q(x) is the quotient, R(x) is the remainder

Since P(x) is already in the form where D(x) = (x+1)(x-3)
Then the remainder is simply P(x) = (x+1)(x-3)Q(x) + a(x+1) + b, where R(x) = [a(x+1) + b]
P(x) = (x+1)(x-3)Q(x) + 3x + 3 - 11
P(x) = (x+1)(x-3)Q(x) + 3x - 8

R(x) = 3x - 8

No 4unit knowledge necessary :/

 

[Trebla]

So as a general rule, when you divide a polynomial P(x) by a non-factor polynomial g(x) of degree n, the degree of R(x) is always n-1? If that makes sense...

No. It is at MOST degree (n - 1)

NB: Here r is used instead to distinguish from the n degree polynomial P(x).
P(x) = A(x)Q(x) + R(x)
When you divide a polynomial P(x) of degree n with a polynomial A(x) of degree r. Assuming that A(x) is not a factor of P(x), then the remainder term R(x) will have degree of AT MOST (r - 1). In other words R(x) can have degree (r - 1) or less.
So for example, if you divide a polynomial of degree 5 with a polynomial of degree 3, your remainder R(x) is either a polynomial of degree 2, 1 or a constant.

The reason for this because the remainder R(x) is irreducible if it has a lower degree than A(x). So for example, with the polynomial P(x) of degree 5, when you divide by A(x) which is of degree 3, gives some random polynomials and some random remainder R(x). If R(x) is of degree 3 or higher, then you can divide R(x) further with A(x) (i.e. R(x) is still divisible by A(x)). As soon R(x), goes below degree 3 to say a quadratic or linear polynomial, you can no longer divide it by A(x).

However, in terms of the question you write the (r - 1) degree polynomial for R(x), because you can't be sure which degree it is (the leading co-efficient may be zero) so you write the (r - 1) degree polynomial without loss of generality.

So if you divide by a quadratic, the remainder has a maximum degree of a linear form ax + b, but there stands the possibility that a = 0 (the remainder is constant) and you can only find it out using the given conditions of the question.

Similarly if divide by a cubic, the remainder has a maximum degree of the form ax² + bx + c, but there stands the possibility that a = 0 (i.e. remainder takes linear form), or both a = b = 0 (i.e. remainder is constant).

 

[akukei]

Oh, yes that makes more sense.

Thanks for clarifying