Question

the polynomial f(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 when f(x) is divided by (x-2).​

Answer 1

Answer:

Given that the equation

f(x) = x4 – 2x3 + 3x2 – ax +b

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

∴ f(-1) = 19 and f(1) = 5

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

⇒ 1 +2 + 3 + a + b = 19

∴ a + b = 13 ——- (1)

According to given condition f(1) = 5

f(x) = x4 – 2x3 + 3x2 – ax

⇒ 14 – 2 3 + 3 2 – a (1) b = 5

⇒ 1 – 2 + 3 – a + b = 5

∴ b – a = 3 —— (2)

solving equations (1) and (2)

a = 5 and b = 8

Now substituting the values of a and b in f(x) , we get

∴ f(x) = x4 – 2x3 + 3x2 – 5x + 8

Also f(x) is divided by (x-2) so remainder will be f(2)

∴ f(x)= x4 – 2x3 + 3x2 – 5x + 8

⇒ f(2) = 16 – 2 × 8 + 3 × 4 – 5 × 2+ 8

= 16–16+12–10+8

= 10

Therefore, f(x) = x4 – 2x3 + 3x2 – ax +b when a=3 and b= 8 is 10.

Answer 2

Answer:

Step-by-step explanation:

We are given these two polynomials. Before that, we would like to apply the substitution method :-

(x-1) -- x-1 = 0

$\implies$ x = 1 ----------------[1]

$\implies$ (x+1) -- x+1 = 0

$\implies$ x = -1 ----------[2]

Now, we will substitute it:-

x⁴-2x³+3x²-ax+b

$\implies$ (1)⁴-2(1)³+3(1)²-a(1)+b = 5

$\implies$ 1 - 2 + 3 -a +b = 5

$\implies$ 2 -a +b = 5

$\implies$ -a+b = 3 ------[3]

$\implies$ (-1)⁴-2(-1)³+3(-1)²-a(-1)+b = 19

$\implies$ 1 +2 +3 +a + b = 19

$\implies$ 6 + a + b = 19

$\implies$ a +b = 13 ----[4]

Now, add it :-

$\implies$ -a + b + a + b = 16

$\implies$ 2b = 16

$\implies$ b = 8.

Now, we will find out a.

$\implies$ a = 13 - 8 [Equation 3]

a = 5

Now, what we will do is take any equation {say 1} and substitute 2 in place of x.

$\implies$ (2)⁴-2(2)³+3(2)²-5(2)+8

$\implies$ 16 - 16 + 12 - 10 + 8

= 10 is the remainder when  f(x) is divided by (x-2).​