5 polynomial pains (1 Viewer)

[Mathman26]

could any1 tell me how to answer these?

 

[gurmies]

Question 1

(i) The divisor has a degree of two. Remainders are always one degree less than the divisor (verify this with any long division example). As a result, the most common form for R(x) is ax + b, in other words - linear.

(ii) P(x) = (x + 1)(x - 4)Q(x) + ax + b (or R(x))

If P(4) = - 5, then 4a + b = - 5 ===> R(4) = - 5

(iii) P(-1) = 5, so - a + b = 5

- 5a = 10 (by subtracting two linear equations)

a = - 2

.'. b = 7

.'. R(x) = - 2x + 7

Question 2

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

= (x - 1)(x + 1)Q(x) + 3x - 1

When divided by (x - 1), P(1) = 2

Therefore, remainder is 2, when P(x) is divided by (x - 1)

Question 5

x^3 - 2x^2 + a = (x + 2)Q(x) + 3

Put x = - 2

(- 2)^3 - 2(- 2)^2 + a = 3

- 8 - 8 + a = 3

a = 19

 

[scardizzle]

Not sure if this is right but for q 3

if P(x) = (x-2)(x-1)Q(x)

then x = 2 and x = 1 are factors hence you sub those values in

and you'll end up with all the terms cancelling out

Question 4

using remainder theorem when x= -1 remainder is -11

therefore 0 + 0 + b = -11

therefore b = -11

using when x = 3

4a + b = 1 sub b = -11

therefore 4a - 11 = 1

therefore a = 3

hence the remainder when P(x) is divided by (x+1)(x-3) = 3(x+1) - 11

= 3x -8