Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 1:

Determine which of the following polynomials has (x + 1) a factor:

(i) x³ + x² + x + 1 (ii) x⁴ + x³ + x² + x + 1

(iii) x⁴ + 3x³ + 3x² + x + 1 (iv) 

Answer:

(i) If (x + 1) is a factor of p(x) = x³ + x² + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x³ + x² + x + 1

p(−1) = (−1)³ + (−1)² + (−1) + 1

= − 1 + 1 − 1 + 1 = 0

Hence, x + 1 is a factor of this polynomial.

(ii) If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x⁴ + x³ + x² + x + 1

p(−1) = (−1)⁴ + (−1)³ + (−1)² + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1

As p(− 1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(iii) If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

p(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1

As p(−1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(iv) If(x + 1) is a factor of polynomial p(x) =  , then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

As p(−1) ≠ 0,

Therefore, (x + 1) is not a factor of this polynomial.

Question 2:

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x³ + x² − 2x − 1, g(x) = x + 1

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

(iii) p(x) = x³ − 4 x² + x + 6, g(x) = x − 3

Answer:

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero.

p(x) = 2x³ + x² − 2x − 1

p(−1) = 2(−1)³ + (−1)² − 2(−1) − 1

= 2(−1) + 1 + 2 − 1 = 0

Hence, g(x) = x + 1 is a factor of the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must

be 0.

p(x) = x³ +3x² + 3x + 1

p(−2) = (−2)³ + 3(−2)² + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As p(−2) ≠ 0,

Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then p(3) must

be 0.

p(x) = x³ − 4 x² + x + 6

p(3) = (3)³ − 4(3)² + 3 + 6

= 27 − 36 + 9 = 0

Hence, g(x) = x − 3 is a factor of the given polynomial.

Question 3:

Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x² + x + k (ii) 

(iii)   (iv) p(x) = kx² − 3x + k

Answer:

If x − 1 is a factor of polynomial p(x), then p(1) must be 0.

(i) p(x) = x² + x + k

p(1) = 0

⇒ (1)² + 1 + k = 0

⇒ 2 + k = 0 ⇒ k = −2

Therefore, the value of k is −2.

(ii) 

p(1) = 0

(iii) 

p(1) = 0

(iv) p(x) = kx² − 3x + k

⇒ p(1) = 0

⇒ k(1)² − 3(1) + k = 0

⇒ k − 3 + k = 0

⇒ 2k − 3 = 0

Therefore, the value of k is .

Question 4:

Factorise:

(i) 12x² − 7x + 1 (ii) 2x² + 7x + 3

(iii) 6x² + 5x − 6 (iv) 3x² − x − 4

Answer:

(i) 12x² − 7x + 1

We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.

Here, 12x² − 7x + 1 = 12x² − 4x − 3x + 1

= 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1)

(ii) 2x² + 7x + 3

We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.

They are p = 6 and q = 1.

Here, 2x² + 7x + 3 = 2x² + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x+ 1)

(iii) 6x² + 5x − 6

We can find two numbers such that pq = −36 and p + q = 5.

They are p = 9 and q = −4.

Here,

6x² + 5x − 6 = 6x² + 9x − 4x − 6

= 3x (2x + 3) − 2 (2x + 3)

= (2x + 3) (3x − 2)

(iv) 3x² − x − 4

We can find two numbers such that pq = 3 × (− 4) = −12

and p + q = −1.

They are p = −4 and q = 3.

Here,

3x² − x − 4 = 3x² − 4x + 3x − 4

= x (3x − 4) + 1 (3x − 4)

= (3x − 4) (x + 1)

Question 5:

Factorise:

(i) x³ − 2x² − x + 2 (ii) x³ + 3x² −9x − 5

(iii) x³ + 13x² + 32x + 20 (iv) 2y³ + y² − 2y − 1

Answer:

(i) Let p(x) = x³ − 2x² − x + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

p(−1) = (−1)³ − 2(−1)² − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (x +1 ) is factor of polynomial p(x).

Let us find the quotient on dividing x³ − 2x² − x + 2 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ x³ − 2x² − x + 2 = (x + 1) (x² − 3x + 2) + 0

= (x + 1) [x² − 2x − x + 2]

= (x + 1) [x (x − 2) − 1 (x − 2)]

= (x + 1) (x − 1) (x − 2)

= (x − 2) (x − 1) (x + 1)

(ii) Let p(x) = x³ − 3x² − 9x − 5

All the factors of 5 have to be considered. These are ±1, ± 5.

By trial method,

p(−1) = (−1)³ − 3(−1)² − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, x + 1 is a factor of this polynomial.

Let us find the quotient on dividing x³ + 3x² − 9x − 5 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ x³ − 3x² − 9x − 5 = (x + 1) (x² − 4x − 5) + 0

= (x + 1) (x² − 5x + x − 5)

= (x + 1) [(x (x − 5) +1 (x − 5)]

= (x + 1) (x − 5) (x + 1)

= (x − 5) (x + 1) (x + 1)

(iii) Let p(x) = x³ + 13x² + 32x + 20

All the factors of 20 have to be considered. Some of them are ±1,

± 2, ± 4, ± 5 ……

By trial method,

p(−1) = (−1)³ + 13(−1)² + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0

As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x).

Let us find the quotient on dividing x³ + 13x² + 32x + 20 by (x + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) + 0

= (x + 1) (x² + 10x + 2x + 20)

= (x + 1) [x (x + 10) + 2 (x + 10)]

= (x + 1) (x + 10) (x + 2)

= (x + 1) (x + 2) (x + 10)

(iv) Let p(y) = 2y³ + y² − 2y − 1

By trial method,

p(1) = 2 ( 1)³ + (1)² − 2( 1) − 1

= 2 + 1 − 2 − 1= 0

Therefore, y − 1 is a factor of this polynomial.

Let us find the quotient on dividing 2y³ + y² − 2y − 1 by y ­− 1.

p(y) = 2y³ + y² − 2y − 1

= (y − 1) (2y² +3y + 1)

= (y − 1) (2y² +2y + y +1)

= (y − 1) [2y (y + 1) + 1 (y + 1)]

= (y − 1) (y + 1) (2y + 1)