Question

Now in each of the pairs of polynomials given below, check whether the first is a factor of the second:

(i) x + 1, x³ − 1

(ii) x − 1, x³ + 1

(iii) x + 1, x³ + 1

(iv) x² − 1, x⁴ − 1

(v) x − 1, x⁴ − 1

(vi) x + 1, x⁴ − 1

(vii) x − 2, x² − 5x + 1

(viii) x + 2, x² + 5x + 6

(ix)

(x) 1.3x − 2.6, x² − 5x + 6

Answer 1

(i)

To find whether (x + 1) is a factor of (x³ − 1), we have to express (x³ − 1) in terms of (x + 1).

It can be done as follows:

(x³ − 1) = (x + 1)(ax² + bx + c) + d

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

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

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

⇒ b = −a = −1

c + b = 0

⇒ c = −b = 1

c + d = −1

⇒ d = −1 − c = −1 − 1 = −2

Thus, when (x³ − 1) is divided by (x + 1), the remainder is −2.

∴ (x + 1) is not a factor of (x³ − 1).

(ii)

To find whether (x − 1) is a factor of (x³ + 1), we have to express (x³ + 1) in terms of (x − 1).

It can be done as follows:

(x³ + 1) = (x − 1)(ax² + bx + c) + d

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

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

For this to hold, we need to equate the coefficients.

We have:

a = 1

b − a = 0

⇒ b = a = 1

c − b = 0

⇒ c = b = 1

−c + d = 1

⇒ d = 1 + c = 1 + 1 = 2

Thus, when (x³ + 1) is divided by (x −1), the remainder is 2.

∴ (x −1) is not a factor of (x³ + 1).

(iii)

To find whether (x + 1) is a factor of (x³ + 1), we have to express (x³ + 1) in terms of (x + 1).

It can be done as follows:

(x³ + 1) = (x + 1)(ax² + bx + c) + d

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

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

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

⇒ b = −a = −1

c + b = 0

⇒ c = −b = 1

c + d = 1

⇒ d = 1 − c = 1 −1 = 0

Thus, when (x³ + 1) is divided by (x +1), the remainder is 0.

∴ (x + 1) is a factor of (x³ + 1).

(iv)

To find whether (x² − 1) is a factor of (x⁴ − 1), we have to express (x⁴ − 1) in terms of (x² − 1).

It can be done as follows:

(x⁴ − 1) = (x² − 1)(ax² + bx + c) + d

⇒(x⁴ − 1) = ax⁴ + bx³ + cx² − ax² − bx − c + d

⇒ (x⁴ − 1) = ax⁴ + bx³ + (c − a)x² − bx − c + d

For this to hold, we need to equate the coefficients.

We have:

a = 1

b = 0

c − a = 0

⇒ c = a = 1

−c + d = −1

⇒ d = −1 + c = −1 + 1 = 0

Thus, when (x⁴ − 1) is divided by (x² − 1), the remainder is 0.

∴ (x² − 1) is a factor of (x⁴ − 1).

(v)

To find whether (x − 1) is a factor of (x⁴ − 1), we have to express (x⁴ − 1) in terms of (x − 1).

It can be done as follows:

(x⁴ − 1) = (x − 1)(ax³ + bx² + cx + d) + e

⇒ (x⁴ − 1) = ax⁴ + bx³ + cx² + dx − ax³ − bx² − cx − d + e

⇒ (x⁴ − 1) = ax⁴ + (b − a)x³ + (c − b)x² + (d − c)x − d + e

For this to hold, we need to equate the coefficients.

We have:

a = 1

b − a = 0

⇒ b = a = 1

c − b = 0

⇒ c = b = 1

d − c = 0

⇒ d = c = 1

−d + e = −1

⇒ e = −1 + d = −1 + 1 = 0

Thus, when (x⁴ − 1) is divided by (x − 1), the remainder is 0.

∴ (x − 1) is a factor of (x⁴ − 1).

(vi)

To find whether (x + 1) is a factor of (x⁴− 1), we have to express (x⁴ − 1) in terms of (x + 1).

It can be done as follows:

(x⁴ −1) = (x + 1)(ax³ + bx² + cx + d) + e

⇒ (x⁴ − 1) = ax⁴ + bx³ + cx² + dx + ax³ + bx² + cx + d + e

⇒ (x⁴ − 1) = ax⁴ + (b + a)x³ + (c + b)x² + (d + c)x + d + e

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

⇒ b = −a = −1

c + b = 0

⇒ c = −b = 1

d + c = 0

⇒ d = −c = −1

d + e = −1

⇒ e = −1 − d = −1 + 1 = 0

Thus, when (x⁴ − 1) is divided by (x + 1), the remainder is 0.

∴ (x + 1) is a factor of (x⁴ − 1).

(vii)

To find whether (x − 2) is a factor of (x² − 5x + 1), we have to express (x² − 5x + 1) in terms of (x − 2).

It can be done as follows:

(x² − 5x + 1) = (x − 2)(ax + b) + c

⇒ (x² − 5x + 1) = ax² + bx − 2ax − 2b + c

⇒ (x² − 5x + 1) = ax² + (b − 2a)x − 2b + c

For this to hold, we need to equate the coefficients.

We have:

a = 1

b − 2a = −5

⇒ b = 2a − 5 = 2 − 5 = −3

−2b + c = 1

⇒ c = 1 + 2b = 1 − 6 = −5

Thus, when (x² − 5x + 1) is divided by (x − 2), the remainder is −5.

∴ (x − 2) is not a factor of (x² − 5x + 1).

(viii)

To find whether (x + 2) is a factor of (x² + 5x + 6), we have to express (x² + 5x + 6) in terms of (x + 2).

It can be done as follows:

(x² + 5x + 6) = (x + 2)(ax + b) + c

⇒ (x² + 5x + 6) = ax² + bx + 2ax + 2b + c

⇒ (x² + 5x + 6) = ax² + (b + 2a)x + 2b + c

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + 2a = 5

⇒ b = −2a +5 = −2 + 5 = 3

2b + c = 6

⇒ c = 6 − 2b = 6 − 6 = 0

Thus, when (x² + 5x + 6) is divided by (x + 2), the remainder is 0.

∴ (x + 2) is a factor of (x² + 5x + 6).

(ix)

To find whether is a factor of (x² − 5x + 6), we have to express (x² − 5x + 6) in terms of .

It can be done as follows:

(x² − 5x + 6) = (ax + b) + c

⇒ (x² − 5x + 6) =

⇒ (x² − 5x + 6)

For this to hold, we need to equate the coefficients.

We have:

⇒ a = 3

= 5

⇒ b = 2a + 15 = 6 + 15 = 21

= 6

⇒ 3c = 18 + 2b = 18 + 42 = 60

Thus, when (x² − 5x + 6) is divided by the remainder is 60.

∴ is not a factor of (x² − 5x + 6).

(x)

To find whether (1.3x − 2.6) is a factor of (x² − 5x + 6), we have to express (x² −5x + 6) in terms of (1.3x + 2.6).

It can be done as follows:

(x² − 5x + 6) = (1.3x + 2.6)(ax + b) + c

⇒ (x² − 5x + 6) = 1.3ax² + 1.3bx + 2.6ax + 2.6b + c

⇒ (x² − 5x + 6) = 1.3ax² + (1.3b + 2.6a)x + 2.6b + c

For this to hold, we need to equate the coefficients.

We have:

1.3a = 1

⇒ a =

1.3b + 2.6a = −5

⇒ 1.3b = −2.6a − 5

= −2 −5

= −7

⇒ b =

2.6b + c = 6

⇒ c = 6 − 2.6b = 6 + 14 = 20

Thus, when (x² − 5x + 6) is divided by (1.3x − 2.6), the remainder is 20.

∴ (1.3x − 2.6) is not a factor of (x² − 5x + 6).