Question

Please do the 14 no. question

Answer 1

The polynomial p(x) = x^4–2x^3+3x^2-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19, respectively. What are a and b?

p(x) = x^4–2x^3+3x^2-ax+b-5 = 0

p(1) = 1–2+3-a+b-5 = 0, or -a+b-3=0, or -a+b = 3 …(1)

p(x) = x^4–2x^3+3x^2-ax+b-19 = 0

p(-1) = 1+2+3+a+b-19=0, or a+b-13=0, or a+b = 13 …(2)

Add (1) and (2), to get

2b=16, or b = 8. From (2), a = 13–8=5

So the polynomial p(x) = x^4–2x^3+3x^2–5x+8=0 is divisible by (x-1) and (x+1) without leaving any remainders.

Check: x^4–2x^3+3x^2–5x+8=0. p(1) = 1–2+3–5+8= 5 as remainder.

x^4–2x^3+3x^2–5x+8=0. p(-1) = 1+2+3+5+8= 19 as remainder.