Polynomial questions from Terry Lee (1 Viewer)

[sinophile]

1. Use long division to find the quotient and remainder of (a^2 -ab +b^2)/(a+b).

2. A polynomial gives remainders of 1 and 4 when it is divided by x+1 and x-3 respectively. What is the remainder when it is divided by (x+1)(x-3)?

3. What is the remainder when x^(99)-99 is divided by x^2)-1? (explain your reasoning)

 

[addikaye03]

1. Use long division to find the quotient and remainder of (a^2 -ab +b^2)/(a+b).

2. A polynomial gives remainders of 1 and 4 when it is divided by x+1 and x-3 respectively. What is the remainder when it is divided by (x+1)(x-3)?

3. What is the remainder when x^(99)-99 is divided by x^2)-1? (explain your reasoning)

2.Let P(x)=Q(x)A(x)+R(x) Where R(x)=ax+b

so when P(x) is divided by (x+1) remainder is 1 ie. P(-1)=(-1+1)(-1-3)+a(-1)+b=1

1=-a+b [1]

When P(x) is divided by (x-3) remainder is 4. ie. P(3)=(4)(0)+3a+b=4 [2]

-a+b=1
3a+b=4

-4a=-3 therefore a=3/4

Sub a into [1]

-3/4+b=1

b=7/4

So When P(x)=(x+1)(x-3)+R(x)

R(x)=3/4(x)+7/4

prettty sure thats right

 

[Trebla]

3. What is the remainder when x^(99)-99 is divided by x^2)-1? (explain your reasoning)

x⁹⁹ - 99 = Q(x)(x² - 1) + R(x)
In this case, R(x) must be linear or constant so to preserve generality let R(x) = ax + b
x⁹⁹ - 99 = Q(x)(x² - 1) + ax + b
Let x = 1
=> a + b = - 98
Let x = -1
=> - a + b = - 100
Hence
b = - 99 and a = 1
So remainder is (x - 99)

 

[sinophile]

2.Let P(x)=Q(x)A(x)+R(x) Where R(x)=ax+b

so when P(x) is divided by (x+1) remainder is 1 ie. P(-1)=(-1+1)(-1-3)+a(-1)+b=1

1=-a+b [1]

When P(x) is divided by (x-3) remainder is 4. ie. P(3)=(4)(0)+3a+b=4 [2]

-a+b=1
3a+b=4

-4a=-3 therefore a=3/4

Sub a into [1]

-3/4+b=1

b=7/4

So When P(x)=(x+1)(x-3)+R(x)

R(x)=3/4(x)+7/4

prettty sure thats right

Thanks.

But doesn't that assume the polynomial P(x) is cubic? It could be any degree for all I know.

 

[Timothy.Siu]

Thanks.

But doesn't that assume the polynomial P(x) is cubic? It could be any degree for all I know.

no
he just left out Q(x)

 

[Trebla]

Thanks.

But doesn't that assume the polynomial P(x) is cubic? It could be any degree for all I know.

The remainder is still linear. When you divide a polynomial of degree p by a polynomial of degree q (p > q), the remainder must have degree less than q. In this case, division of a quadratic polynomial gives either a linear or constant polynomial as the remainder. So to maintain generality, we let the remainder be linear because if it were constant the co-efficient of x would be found to be 0.

 

[addikaye03]

no
he just left out Q(x)

Ahh shit, yer sorry... add Q(x) to the factors...answer and everything is still the same though