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

Solution:

The zeroes of a polynomial are values of x in a polynomial that makes the polynomial equal to zero.

Given that f (x) = x⁴ - 2x³ + 3x² - ax + b divided by x - 1 and x + 1 leaves remainder 5 and 19.

⇒ f(1) = 5 and f(-1) = 19

So,  (1)⁴ - 2 (1)³ + 3 (1)² - a(1) + b = 5

⇒ 1 - 2 + 3 - a + b = 5

⇒ - a + b = 5 - 2

⇒ - a + b = 3 ----- (1)

f (-1) = 19

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

⇒1 + 2 + 3 + a + b = 19

⇒ a + b = 19 - 6 

⇒ a + b = 13 ----- (2)

By adding equation (1) and (2), we get

2b = 16 ⇒ b = 8

⇒ a = 13 - 8 = 5   (by substituting value of b in equation (2))

Hence, f(x) = x⁴ - 2x³ + 3x² - 5x + 8

It is also given that f (x) = x⁴ - 2x³ + 3x² - ax + b is divided by (x - 3).

Therefore, f(3)  = (3)⁴ - 2 (3)³ + 3 (3)² - 5 (3) + 8

⇒ f(3) = 81 - 54 + 27 - 15 + 8

⇒ f(3) = 47

Thus, the value of remainder when f (x) = x⁴ - 2x³ + 3x² - ax + b is divided by (x - 3) is 47.

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

Summary:

The remainder when f(x) = x⁴ - 2x³ + 3x² - ax + b  is divided by x - 3 is 47