Polynomials: 9 Math

NCERT Exercise 2.2: 9th math

Important points

(i) A zero of a polynomial need not be 0

(ii) 0 may be a zero of a polynomial

(iii) Every linear polynomial has one and only one zero

(iv) A polynomial can have more than one zero

NCERT Exercise 2.2 Polynomial Question (1) Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0

Answer

Given, x = 0

And, p(x) = 5x – 4x² + 3

Thus, p(0) = 5 × 0 – 4 × 0² + 3

= 0 – 4 × 0 + 3

= 0 – 0 + 3

= 3

Thus, p(0) = 3 Answer

(ii) x = –1

Answer

Given, p(x) = 5x – 4x² + 3

Thus, p(–1) = 5 (–1) – 4 (–1)² + 3

= –5 – 4 × 1 + 3

= –5 – 4 + 3

= –9 + 3

= –6

Thus, p(–1) = – 6 Answer

(iii) x = 2

Answer

Given, p(x) = 5x – 4x² + 3

Thus, p(2) = 5 × 2 – 4 × (2)² + 3

= 10 – 4 × 4 + 3

= 10 – 16 + 3

= – 6 + 3

= –3

Thus, p(2) = –3 Answer

NCERT Exercise 2.2 Polynomials class ninth math Question (2) Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y² – y + 1

Answer

Given, p(y) = y² – y + 1

Thus, P(0) = 0² – 0 + 1

= 0 – 0 + 1

= 1

Thus, p(0) = 1

And, p(1) = 1² – 1 + 1

= 1 – 1 + 1

= 1

Thus, p(1) = 1

And, p(2) = 2² – 2 + 1

= 4 – 2 + 1

= 2 + 1 = 3

Thus, p(2) = 3

Thus, p(0) = 1, p(1) = 1 and p(2) = 3 Answer

(ii) p(t) = 2 + t + 2t² – t³

Answer

Given, p(t) = 2 + t + 2t² – t³

Thus, p(0) = 2 + 0 + 2 (0)² – (0)³

= 2 + 2 × 0 – 0

= 2 + 0 – 0

= 2

Thus, p(0) = 2

And, p(1) = 2 + 1 + 2 (1)² – (1)³

= 2 + 1 + 2 × 1 – 1

= 2 + 1 + 2 – 1

= 5 – 1

= 4

Thus, p(1) = 4

And, p(2) = 2 + 2 + 2 (2)² – (2)³

= 4 + 2 × 4 – 8

= 4 + 8 – 8

= 4

Thus, p(2) = 4

Thus, p(0) = 2, P(1) = 4 and p(2) = 4 Answer

(iii) p(x) = x³

Answer

Given, p(x) = x³

Thus, p(0) = 0³

⇒ p(0) = 0

And, p(1) = 1³

⇒ p(1) = 1

And, p(2) = 2³

⇒ p(2) = 8

Thus, p(0) = 0, p(1) = 1 and p(2) = 8 Answer

(iv) p(x) = (x – 1)(x + 1)

Answer

Given, p(x) = (x – 1)(x + 1)

Thus, p(0) = (0 – 1)(0 + 1)

= –1 × 1

= –1

Thus, p(0) = –1

And, p(1) = (1 – 1) (1 + 1)

= 0 × 2 = 0

Thus, p(1) = 0

And, p(2) = (2 – 1) (2 + 1)

= 1 × 3 = 3

Thus, p(2) = 3

Thus, p(0) = –1, p(1) = 0 and p(2) = 3 Answer

NCERT Exercise 2.2 Polynomials class ninth math Question (3) Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x=–1/3

Answer

Given, p(x) = 3x + 1

Thus, `p(-1/3)=3×-1/3 + 1`

= –1 + 1 = 0

Thus, `p(-1/3) = 0`

Thus, `x =-1/3` is the zero of given polynomial. Answer

(ii) `p(x) = 5x-pi, x=4/5`

Answer

Given, `p(x) = 5x-pi`

Thus, `p(4/5) = 5xx4/5-pi`

`=4-22/7`

`=(28-22)/7`

`=>p(4/5)=6/7`

Thus, `4/5` is not the zero of given polynomial.

(iii) p(x) = x² –1, x = 1, –1

Answer

Given, p(x) = x² –1,

Thus, p(1) = 1² – 1

= 1 – 1 =0

⇒p(1) = 0

And, p(–1) = (-1)² – 1

= 1 – 1 = 0

⇒ p(–1) = 0

Thus, 1 and –1 both are zeroes of the given polynomial Answer

(iv) p(x) = (x + 1) (x – 2), x = –1, 2

Answer

Given, p(x) = (x + 1) (x – 2)

Thus, p(–1) = (–1 + 1)(–1 – 2)

= 0 ×(–3)

⇒ p(–1) = 0

And, p(2) = (2 + 1)(–2 – 2)

= 3 × 0

⇒ p(2) = 0

Thus, only –1 and 2 are the zeroes of given polynomial Answer

(v) p(x) = x², x = 0

Answer

Given, p(x) = x²

Thus, p(0) = 0²

⇒ p(0) = 0

Thus, 0 is the zero of given polynomial Answer

(vi) `p(x) = lx+m, x =-m/l`

Answer

Given, `p(x) = lx+m`

Thus, `p(-3/l) = lxx-m/l+m`

= – m + m

`=>p(-m/l) = 0`

Thus, `-m/l` is the zero of the given polynomial Answer

(vii) p(x) = 3x² – 1, `x=-1/sqrt3, 2/sqrt2`

Answer

Given, p(x) = 3x² – 1

Thus, `p(-1/sqrt3) = 3xx(-1/sqrt3)^2 – 1`

`=3xx1/3 – 1`

= 1 – 1

`=>p(-1/sqrt3)=0`

And, `p(2/sqrt3) = 3xx(2/sqrt3)^2-1`

`=3xx4/3-1`

= 4 – 1

`=>p(2/sqrt3) = 3

Thus, only `–1/sqrt3` is the zero of given polynomial Answer

(viii) p(x) = 2x + 1, x = 1/2

Answer

Given, p(x) = 2x + 1

Thus, `p(1/2) =2xx1/2+1`

= 1 + 1

`=> p(1/2)=2`

Thus, 1/2 is not the zero of given polynomial Answer

NCERT Exercise 2.2 Polynomials class ninth math Question (4) Find the zero of the polynomial in each of the following cases

(i) p(x) = x + 5

Answer

Given, p(x) = x + 5

Thus, p(x) = 0

⇒ x + 5 = 0

Thus, x = – 5

Thus, –5 is the zero of the given polynomial Answer

(ii) p(x) = x – 5

Answer

p(x) = x – 5

Thus, p(x) = 0

⇒ x – 5 = 0

⇒ x = 5

Thus, 5 is the zero of the given polynomial Answer

(iii) p(x) = 2x + 5

Answer

Given, p(x) = 2x + 5

Thus, p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = – 5

`=> x = 5/2`

Thus, `5/2` is the zero of the given polynomial Answer

(iv) p(x) = 3x – 2

Answer

Given, p(x) = 3x – 2

Thus, p(x) = 0

⇒ 3x – 2 = 0

⇒ 3x = 2

`=>x=2/3`

Thus, `2/3` is the zero of the given polynomial Answer

(v) p(x) = 3x

Answer

Given, p(x) = 3x

Thus, p(x) = 0

⇒ 3x = 0

`=> x = 0/3 = 0`

Thus, 0 is the zero of the given polynomial Answer

(vi) p(x) = ax, a ≠0

Answer

Given, p(x) = ax

Thus, p(x) = 0

⇒ ax = 0

`=>x = 0/a`

⇒ x = 0

Thus, 0 is the zero of the given polynomial Answer

(vii) p(x) = cx + d, c ≠ 0, c, and d are real numbers

Answer

Given, p(x) = cx + d

Thus, when p(x) = 0

⇒ c x + d = 0

⇒ c x = – d

`=> x = -d/x`

Thus, `-d/x` is the zero of the given polynomial Answer

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