Question

Example 3. On dividing
f(x) = x⁴ - 2x³ + 3x² - ax + b by (x - 1) and (x + 1),
we get remainder 5 and 19 respectively. Find the
remainder when fix) is divided by (x - 2).​

Answer 1

Step-by-step explanation:

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 + b

⇒ 1⁴ – 2*1³ + 3*1² – 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) = 2⁴-2*2³+3*2²-5*2+8

⇒ f(2) = 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:

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-3) so remainder will be f(3)

∴ f(x)= x4 – 2x3 + 3x2 – 5x + 8

⇒ f(3) = 34 – 2 × 33 + 3 × 32 – 5 × 3 + 8

= 81 – 54 + 27 – 15 + 8

= 47

Therefore, f(x) = x4 – 2x3 + 3x2 – ax +b when a=3 and b= 8 is 47