Question

Now in each of the following, find the quotient and remainder on dividing p(x) by q(x) and write then in the form p(x) = u(x) q(x) + v(x).

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Answer 1

(i)

Given: p(x) = 6x² + 7x + 1, q(x) = 3x − 1

When 6x² + 7x + 1 is divided by 3x − 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and the remainder be c.

⇒ (ax + b)(3x − 1) + c = 6x² + 7x + 1

⇒ 3ax² − ax + 3bx − b + c = 6x² + 7x + 1

⇒ 3ax² + (−a + 3b) x + (−b + c) = 6x² + 7x + 1

On comparing the coefficients, we get:

3a = 6, −a + 3b = 7 and −b + c = 1

3a = 6

⇒ a = 2

−a + 3b = 7

⇒ −2 + 3b = 7

⇒ 3b = 7 + 2

⇒ 3b = 9

⇒ b = 3

−b + c = 1

⇒ −3 + c = 1

⇒ c = 1 + 3 = 4

Hence, the quotient is 2x + 3 and the remainder is 4.

Thus, u(x) = 2x + 3 and v(x) = 4.

p(x) = u(x) q(x) + v(x)

⇒ 6x² + 7x + 1 = (2x + 3)(3x − 1) + 4

(ii)

Given: p(x) = 8x² − 18x − 5, q(x) = 2x − 5

When 8x² − 18x − 5 is divided by 2x − 5, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and remainder be c.

⇒ (ax + b)(2x − 5) + c = 8x² − 18x − 5

⇒ 2ax² − 5ax + 2bx − 5b + c = 8x² − 18x − 5

⇒ 2ax² + (−5a + 2b)x + (−5b + c) = 8x² − 18x − 5

On comparing the coefficients, we get:

2a = 8, − 5a + 2b = −18 and −5b + c = −5

2a = 8

⇒ a = 4

−5a + 2b = −18

⇒ −5(4) + 2b = −18

⇒ −20 + 2b = −18

⇒ 2b = −18 + 20

⇒ 2b = 2

⇒ b = 1

−5b + c = −5

⇒ −5(1) + c = −5

⇒ −5 + c = −5

⇒ c = −5 + 5 = 0

Hence, the quotient is 4x + 1 and the remainder is 0.

Thus, u(x) = 4x + 1 and v(x) = 0.

p(x) = u(x) q(x) + v(x)

⇒ 8x² − 18x − 5 = (4x + 1) (2x − 5) + 0

(iii)

Given: p(x) = x² − 2, q(x) = 2x − 4

When x² − 2 is divided by 2x − 4, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and the remainder be c.

⇒ (ax + b) (2x − 4) + c = x² − 2 … (1)

⇒ 2ax² − 4ax + 2bx − 4b + c = x² − 2

⇒ 2ax² + (−4a + 2b) x + (−4b + c) = x² − 2

On comparing the coefficients, we get:

2a = 1, −4a + 2b = 0 and −4b + c = −2

2a = 1

⇒ a =

−4a + 2b = 0

−4b + c = −2

⇒ −4(1) + c = −2

⇒ −4 + c = −2

⇒ c = − 2 + 4 = 2

Hence, the quotient is and the remainder is 2.

Thus, u(x) = and v(x) = 2

p(x) = u(x) q(x) + v(x)

x² − 2 = (2x − 4) + 2

(iv)

Given: p(x) = 4x² + 3x + 1, q(x) = 4x + 1

When 4x² + 3x + 1 is divided by 4x + 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and remainder be c.

⇒ (ax + b)(4x + 1) + c = 4x² + 3x + 1

⇒ 4ax² + ax + 4bx + b + c = 4x² + 3x + 1

⇒ 4ax² + (a + 4b)x + (b + c) = 4x² + 3x + 1

On comparing the coefficients, we get:

4a = 4, a + 4b = 3 and b + c = 1

4a = 4

⇒ a = 1

a + 4b = 3

⇒ 1 + 4b = 3

⇒ 4b = 3 − 1

⇒ 4b = 2

⇒ b =

b + c = 1

⇒ + c = 1

⇒ c = 1 − =

Hence, the quotient is x +and the remainder is .

Thus, u(x) = x + and v(x) =

p(x) = u(x) q(x) + v(x)

4x² + 3x + 1 = (4x + 1) +

(v)

Given: p(x) = x³ + x² − 3x + 2, q(x) = x − 1

When x³ + x² − 3x + 2 is divided by x − 1, the quotient must be a polynomial of degree 2 and the remainder must be a constant.

Let the quotient be ax² + bx + c and the remainder be d.

⇒ (ax² + bx + c)(x − 1) + d = x³ + x² − 3x + 2

⇒ ax³ − ax² + bx² − bx + cx − c + d = x³ + x² − 3x + 2

⇒ ax³ + (−a + b)x² + (−b + c)x + (−c + d) = x³ + x² − 3x + 2

On comparing the coefficients, we get:

a = 1, −a + b = 1, −b + c = −3 and −c + d = 2

− a + b = 1

⇒ −1 + b = 1

⇒ b = 1 + 1 = 2

− b + c = −3

⇒ −2 + c = −3

⇒ c = −3 + 2 = −1

− c + d = 2

⇒ −(−1) + d = 2

⇒ 1 + d = 2

⇒ d = 2 − 1 = 1

Hence, the quotient is x² + 2x − 1 and the remainder is 1.

Thus, u(x) = x² + 2x − 1 and v(x) = 1.

p(x) = u(x) q(x) + v(x)

⇒ x³ + x² − 3x + 2 = (x² + 2x − 1)(x − 1) + 1

(vi)

Given: p(x) = x³ − 1, q(x) = x − 1

When x³ − 1 is divided by x − 1, the quotient must be a polynomial of degree 2 and the remainder must be a constant.

Let the quotient be ax² + bx + c and the remainder be d.

⇒ (ax² + bx + c)(x − 1) + d = x³ − 1

⇒ ax³ − ax² + bx² − bx + cx − c + d = x³ − 1

⇒ ax³ + (−a + b)x² + (−b + c)x + (−c + d) = x³ − 1

On comparing the coefficients, we get:

a = 1, −a + b = 0, −b + c = 0 and −c + d = −1

−a + b = 0

⇒ −1 + b = 0

⇒ b = 0 + 1 = 1

−b + c = 0

⇒ −1 + c = 0

⇒ c = 0 + 1 = 1

−c + d = −1

⇒ −1 + d = −1

⇒ d = −1 + 1 = 0

Hence, the quotient is x² + x + 1 and the remainder is 0.

Thus, u(x) = x² + x + 1 and v(x) = 0

p(x) = u(x) q(x) + v(x)

x³ − 1 = (x² + x + 1)(x − 1) + 0

(vii)

Given: p(x) = x³ + 2x² + 4x + 2, q(x) = x² + x + 1

When x³ + 2x² + 4x + 2 is divided by x² + x + 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant or a polynomial of degree 1.

Let the quotient be ax + b and the remainder be cx + d.

⇒ (ax + b)(x² + x + 1) + (cx + d) = x³ + 2x² + 4x + 2

⇒ (ax + b)(x² + x + 1) + (cx + d) = x³ + 2x² + 4x + 2

⇒ ax³ + ax² + ax + bx² + bx + b + cx + d = x³ + 2x² + 4x + 2

⇒ ax³ + (a + b)x² + (a + b + c)x + (b + d) = x³ + 2x² + 4x + 2

On comparing the coefficients, we get:

a = 1, a + b = 2, a + b + c = 4 and b + d = 2

a + b = 2

⇒ 1 + b = 2

⇒ b = 2 − 1 = 1

a + b + c = 4

⇒ 1 + 1 + c = 4

⇒ c = 4 − 2 = 2

b + d = 2

⇒ 1 + d = 2

⇒ d = 2 − 1 = 1

Hence, the quotient is x + 1 and the remainder is 2x + 1.

Thus, u(x) = x + 1 and v(x) = 2x + 1.

p(x) = u(x) q(x) + v(x)

⇒ x³ + 2x² + 4x + 2 = (x + 1)(x² + x + 1) + (2x + 1)