1993 HSC - Polynomials. (1 Viewer)

[Divinity_]

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?

 

[ninetypercent]

think about it. x^2 - 1 = (x+1)(x-1)

so when P(x) is divided by (x+1)(x-1) the remainder is 3x -1

and now P(x) is divided by (x-1) - one of the factors of x^2 -1..

so what should the remainder be then?

 

[Divinity_]

That's what I got but BOB said 2 :S ?

 

[12em12]

That's what I got but BOB said 2 :S ?

ill answer it x^2-1 = (x-1)(x+)

general form of polynomial is:

p(x)= Q(X)A(X) + R(X)

WHERE Q(X) = QUOTENT A(X)= ANSWER R(X)= REMAINDER NOT SURE IF THEY ARE EXACT TERMS BUT YOU UNDERSTAND WHAT I MEAN

NOW P(X) = (X-1)(X+1)A(x) + R(X)
NOW R(X) = 3X -1

P(1) SHOULD GIVE THE ANSWER FOR THE REMAINDER WHEN P(X) DIVIDED BY X-1 RIGHT, SO SUB X=1

P(1)= (1-1)(1+1)A(x) + 3(1)-1
HENCE P(1) = 2

THATS UR ANSWER

 

[tommykins]

the actual polynomial is P(x) = Q(x)(x^2-1) + R(x), not P(x) = (x^2-1)+ R(X)

 

[12em12]

the actual polynomial is P(x) = Q(x)(x^2-1) + R(x), not P(x) = (x^2-1)+ R(X)

yea i understand that but scince it equals to zero when u sub x=1 it does not matter and i did uses the form p(x)= A(x)Q(x) + R(x)

i just forgot to add Q(x) in my other steps so thnx for correcting

 

[tommykins]

yeah essentially it doesnt matter, but just on a technicality