Question

The polynomial p(x) =x⁴-2x³+3x²-ax+b when divided by (x-1) and (x+1) leaves the remainder 5 and 19 respectively . Find the values of a and b .Hence find the remainder​

Answer 1

Answer:

Step-by-step explanation:

p(x)=x^4–2x^3+3x^2-ax+b

By remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.

By the given condition, we have

p(1)=5 and p(-1)=19

=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19

=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19

=> -a+b=5–1+2–3 and 1+2+3+a+b=19

=> -a+b=3 and a+b=19–1–2–3

=> -a+b=3 and a+b=13

Adding these two equations,we get

-a+b+a+b=3+13

=> 2b=16

=> 2b/2=16/2

=> b=8

Putting b=8 in a+b=13 , we get

a+8=13

=> a=13–8

=> a=5

Therefore, a=5 and b=8 .

Answer 2

Answer:

when p(x) is divided by (x+1) and (x-1), the remainders are 19 and 5 respectively.

therefore, p(-1)= 19 and p(1)= 5

⇒(-1)⁴- 2(-1)³ + 3(-1)² - a(-1) + b = 19

⇒1 + 2 + 3 + a + b = 19

therefore, a + b = 19 - 6 = 13               [1]

Again, p(1)= 5

⇒(1)⁴ - 2(1)³ + 3(1)² - a(1) + b = 5

⇒1 - 2 + 3 - a + b = 5

therefore, b - a = 5 - 2 = 3               [2]

Solving equation [1] and [2], we get

therefore, a = 5 and b = 8

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