Question

If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)

Answer 1

P(x) = ax² + bx + c.

When P(x) is divided by x + 3, x – 5 and x – 1.

The remainders are respectively P(– 3), P(5) and P(1).

We are given that P(– 3) = 21

P(5) = 61

P(1) = 9

Now P(– 3) = 21

⇒ a(– 3)² + b(– 3) + c = 21

⇒ 9a – 3b + c = 21  ........(1)

P(5) = 61

⇒ a(5)² + b(5) + c = 61

⇒ 25a + 5b + c = 61  .......(2)

P(1) = 9

⇒ a(1)² + b(1) + c = 9

⇒ a + b + c = 9   .......(3)

Now the matrix form of the above three equations is

`[(9, -3, 1),(25, 5, 1),(1, 1, 1)] [("a"),("b"),("C")] = [(21),(61),(9)]`

(i.e) AX = B

The augmented matrix (A, B) is

[A, B] = `[(9, -3, 1, 21),(25, 5, 1, 61),(1, 1, 1, 9)]`

`˜ [(1, 1, 1, 9),(25, 5, 1, 61),(9, -3, 1, 21)] "R"_1 ↔ "R"_3`

`[(1, 1, 1, 9),(25, 5, 1, 61),(9, -3, 1, 21)] ˜ [(1, 1, 1, 9),(0, -20, -24, -164),(0, -12, -8, -60)] {:("R"_2 -> "R"_2 - 25"R"_1),("R"_3 -> "R"_3 - 9"R"_1):}`

`˜ [(1, 1, 1, 9),(0, -20, -24, -164),(0, 12, 8, 60)] "R"_3 -> - "R"_3`

`˜ [(1, 1, 1, 9),(0, -20, -24, -164),(0, 0, -32, -192)] "R"_3 -> - 5"R"_3 + 3"R"_2`

The above matrix is in echelon form now writing the equivalent equations.

`[(1, 1, 1),(0, -20, -24),(0, 0, -32)][("a"),("b"),("c")] = [(9),(-164),(-192)]`

(i.e) a + b + c = 9

– 20b – 24c = – 164

– 32c = – 192

From (3)

⇒ c = `(- 192)/(- 32)` = 6

Substituting c = 6 in (2) we get

– 20b – 24(6) = – 164

⇒ – 20b = – 164 + 144 = – 20

⇒ b = 1

Substituting c = 6, b = 1 in (1) we get

a + 1 + 6 = 9

⇒ a = 9 – 7 = 2

So a = 2, b = 1, c = 6