Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.

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

(i) x^{3} + x^{2} + x + 1

(ii) x^{4} + x^{3} + x^{2} + x + 1

(iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

(iv) x^{3} – x^{2} – (2 + \( \sqrt{2} \))x + \( \sqrt{2} \)

Answer:

(i) If (x + 1) is a factor of p(x)

=x^{3} + x^{2} + x + 1, p(-1) must be zero.

p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

= – 1 + 1 – 1 + 1 = 0

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

(ii) If (x+ 1) is a factor of p(x)

= x^{4} + x^{3} + x^{2} + x + 1, p(-1) must be zero.

∴ p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1

= 1 – 1 + 1 – 1 + 1 = 1 ≠ 0

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

(iii) If (x + 1) is a factor of polynomial

x^{4} + 3x^{3} + 3x^{2} + x + 1, p(-1) must be 0.

P(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1

= 1 – 3 + 3 – 1 + 1 = 1 ≠ 0

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

(iv) If (x + 1) is a factor of polynomial

p(x) = x^{3} – x^{2} – (2 + \( \sqrt{2} \))x + \( \sqrt{2} \) , p(-1) must be 0.

p(-1) = (-1)^{3} – (-1)^{2} – (2 + \( \sqrt{2} \))(-1) + \( \sqrt{2} \)

= -1 – 1 + 2 + \( \sqrt{2} \) + \( \sqrt{2} \)

= 2\( \sqrt{2} \) ≠ 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) = 2x^{3} + x^{2} – 2x – 1, g(x) = x + 1

(ii) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

(iii) p(x) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

Answer:

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

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 p(x).

(ii) If g(x) = x + 2 is a factor of given polynomial p(x),then p(-2) must be 0.

p(-2) = (-2)^{3} + 3(-2)^{2} + 3(-2) + 1

= – 8 + 12 – 6 + 1 = -1 ≠ 0

Hence g(x) = x + 2 is not a factor of p(x).

(iii) If g(x) = x – 3 is a factor of given polynomial p(x), then p(3) must be 0.

p(3) = (3)^{3} – 4(3)^{2} + 3 + 6 = 27 – 36 + 9 = 0

Therefore, g(x) = x – 3 is a factor of p(x).

Page – 44

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) = x^{2} + x + k

(ii) p(x) = 2x^{2} + kx + \( \sqrt{2} \)

(iii) p(x) = kx^{2} – \( \sqrt{2} \) x + 1

(iv) p(x) = kx^{2} – 3x + k

Answer:

(i) If x – 1 is a factor of polynomial p(x)

= x^{2} + x + k, then p(1) = 0

⇒ (1)^{2}+ 1 + k = 0

⇒ 2 + k = 0

⇒ k = – 2

Therefore, value of k is – 2.

(ii) If x – 1 is a factor of polynomial p(x) = 2x^{2} + kx + \( \sqrt{2} \) , then p(1) = 0

⇒ 2(1)^{2} + k(1) + = 0

⇒ 2 + k + \( \sqrt{2} \) = 0

⇒ k = – 2 – \( \sqrt{2} \) = – (2 + \( \sqrt{2} \))

Therefore, value of k is – (2 + \( \sqrt{2} \) ).

(iii) If x – 1 is a factor of polynomial p(x) = kx^{2} – \( \sqrt{2} \) x + 1, then p(1) = 0

⇒ k(1)^{2} – \( \sqrt{2} \) (1) + 1 = 0

⇒ k – \( \sqrt{2} \) +1 = 0

⇒ k = \( \sqrt{2} \) – 1

Therefore, value of k is \( \sqrt{2} \) – 1.

(iv) If x – 1 is a factor of polynomial p(x) = kx^{2} – 3x + k, then p(1) = 0

⇒ k(1)^{2} – 3(1) + k = 0

⇒ k – 3 + k= 0

⇒ 2k – 3 = 0

Therefore, value of k is \(\frac{3}{2}\)

Question 4.

Factorise:

(i) 12x^{2} – 7x + 1

(ii) 2x^{2} + 7x + 3

(iii) 6x^{2} + 5x – 6

(iv) 3x^{2} – x – 4

Answer:

(i) 12x^{2} – 7x + 1

= 12x^{2} – 4x – 3x+ 1

= 4x (3x – 1) – 1 (3x – 1)

= (3x – 1) (4x – 1)

(ii) 2x^{2} + 7x + 3

= 2x^{2} + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x + 1)

(iii) 6x^{2} + 5x – 6

= 6x^{2} + 9x – 4x – 6

= 3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

(iv) 3x^{2} – x – 4

= 3x^{2} – 4x + 3x – 4

= x (3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

Question 5.

Factorise:

(i) x^{3} – 2x^{2} – x + 2

(ii) x^{3} – 3x^{2} – 9x – 5

(iii) x^{3} + 13x^{2} + 32x + 20

(iv) 2y^{3} + y^{2} – 2y – 1

Answer:

(i) Let p(x) = x^{3} – 2x^{2} – x + 2

Factors of 2 are ±1 and ± 2

By trial method, we find that p(-1) = 0

So, (x + 1) is factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

= (x + 1) (x^{2} – 3x + 2)

= (x + 1) (x^{2} – x – 2x + 2)

= (x+ 1) {x(x – 1) – 2(x – 1)}

= (x + 1) (x – 1) (x – 2)

(ii) Let p(x) = x^{3} – 3x^{2} – 9x – 5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x – 5) is factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

= (x – 5) (x^{2} + 2x + 1)

= (x – 5) (x^{2} + x + x + 1)

= (x – 5) {x(x + 1) +l(x + 1)}

= (x – 5) (x + 1) (x + 1)

(iii) Let p(x) = x^{3} + 13x^{2} + 32x + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

P(-1) = 0

So, (x+1) is factor ofp(x)

Now, Dividend = Divisor × Quotient + Remainder

= (x + 1) (x^{2} + 12x + 20)

= (x + 1) (x^{2} + 2x + 10x + 20)

= (x + 1) {x(x+2) +10(x+2)}

= (x + 1) (x+2) (x+10)

(iv) Let p(y) = 2y^{3} + y^{2} – 2y – 1

Factors of ab = 2 × (-1) = – 2 are ±1 and ±2

By trial method, we find that p(1) = 0

So, (y – 1) is factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder

= (y – 1) (2y^{2} + 3y + 1)

= (y – 1) (2y^{2} + 2y + y + 1)

= (y – 1) {2y(y + 1) + 1(y + 1)}

= (y – 1 )(2y + 1) (y + 1)