Question

If p(x) = x^4 - 2x^3 + 3x^2 - ax + b is a polynomial such that when it is divided by (x-1) and (x+1),
the remainder are respectively 6 and 14. Determine the remainder when p(x) is divided by (x-2).​

Answer 1

Answer:-

Given Polynomial : x⁴ - 2x³ + 3x² - ax + b

It is also given that,

If the polynomial is divided by (x - 1) & (x + 1) the remainders obtained are 6 , 14.

So,

P(x) is divided by (x - 1).

⟶ x - 1 = 0

⟶ x = 1

Hence,

⟶ P(1) = (1)⁴ - 2(1)³ + 3(1)² - a(1) + b

⟶ 6 = 1 - 2 * 1 + 3*1 - a + b

⟶ 6 - 1 + 2 - 3 = b - a

⟶ 4 = b - a -- equation (1)

Similarly,

P(x) is divided by (x + 1).

⟶ x + 1 = 0

⟶ x = - 1

Hence,

⟶ P( - 1) = (- 1)⁴ - 2( - 1)³ + 3( - 1)² - a( - 1) + b

⟶ 14 = 1 - 2 ( - 1) + 3 * 1 + a + b

⟶ 14 - 1 - 2 - 3 = a + b

⟶ 8 = a + b -- equation (2)

Add equations (1) & (2).

⟶ b - a + a + b = 4 + 8

⟶ 2b = 12

⟶ b = 12/2

⟶ b = 6

Substitute b value in equation (1).

⟶ 6 - a = 4

⟶ 6 - 4 = a

⟶ 2 = a

Now,

P(x) is divided by (x - 2).

⟶ x - 2 = 0

⟶ x = 2

Hence,

⟶ P(2) = (2)⁴ - 2(2)³ + 3(2)² - 2(2) + 6

[ Putting the values of a and b too ]

⟶ P(2) = 16 - 2 * 8 + 3(4) - 4 + 6

⟶ P(2) = 16 - 16 + 12 - 4 + 6

⟶ P(2) = 14

Therefore, when P(x) is divided by (x - 2) , the remainder obtained is 14.

Answer 2

Step-by-step explanation:

p(x)=x⁴-2x³+3x²-ax+b

in 1st case

g(x)=x-1

as g(x)=0

x-1=0

x=1

so acc. to remainder theorem

R=p(1)

R=1⁴-2×1³+3×1²-a×1+b

6=2-a+b

b-a=4 eq.1

second case

f(x)=x+1

as f(x)=0

x+1=0

x=-1

acc.to remainder theorem

R=p(-1)

14=(-1)⁴-2×(-1)³+3×(-1)²-a×(-1)+b

14=1+2+3+a+b

b+a=8 ...eq.2

eq.1+eq.2

b-a+b+a=8+4

2b=12

b=6

after substituting in any equations

a comes to be 2

so p(x)=x⁴-2x³+3x²-2x+6

so p(2)=16-16+12-4+6

2 is remainder