**Find the value of the polynomial 5x – 4x**^{2}+ 3 at

**(i) x = 0**

**(ii) x = -1**

**(iii) x = 2**

Solution:

P(x) = 5x – 4x^{2} + 3

(i) x = 0 , P(0) = 5(0) – 4(0)^{2} + 3 = 0 – 0 + 3 = 3

(ii) x = -1, P(-1) = 5(-1) – 4(-1)^{2} + 3 = -5 – 4 + 3 = -9 + 3 = -6

(iii) x = 2, P(2)= 5(2)– 4(2)^{2} + 3 = 10 – 16 + 3 = – 3

**Find p(0), p(1) and p(2) for each of the following polynomials:**

**(i) p(y) = y ^{2} – y + 1**

**(ii) p(t) =2 + t + 2t ^{2} – t^{3}**

**(iii) p(x) = x ^{3}**

**(iv) p(x) = (x – 1)(x + 1)**

Solution:

(i) p(y) = y^{2} – y + 1

p(0) = 0^{2} – 0 + 1 = 1

p(1) = 1^{2} – 1 + 1 = 1

p(2) = 2^{2} – 2 + 1 = 4 – 2 + 1 = 3

(ii) p(t) =2 + t + 2t^{2} – t^{3}

p(0) = 2 + (0) + 2(0)^{2} – (0)^{3} = 2

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

p(2) = 2 + (2) + 2(2)^{2} – (2)^{3} = 2 + 2 + 8 – 8 = 4

(iii) p(x) = x^{3}

p(0) = 0

p(1) = 1^{3} = 1

p(2) = 2^{3} = 8

(iv) p(x) = (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = -1 x 1 = -1

p(1) = (1 – 1)(1 + 1) = 0x2 = 0

p(2) = (2 – 1)(2 + 1) = 1 x 3 = 3

**Verify whether the following are zeroes of the polynomial, indicated against them**

**(i) p(x) = 3x + 1 , x = – ^{1}/_{3}**

**(ii) p(x) = 5x – π , x = ^{4}/_{5}**

**(iii) p(x) = x ^{2} – 1 , x = 1, – 1**

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

**(v) p(x) = x ^{2} , x = 0**

**(vi) p(x) = lx + m , x = – ^{m}/_{l}**

**(vii) p(x) = 3x ^{2} – 1 , x = – ^{1}/_{√3}, ^{2}/_{√3}**

**(viii) p(x) = 2x + 1, x = ^{1}/_{2}**

Solution:

(i) p(x) = 3x + 1 , x = –^{1}/_{3}

Yes zeroes of the polynomial. 3x + 1 = 0 for x = –^{1}/_{3}

(ii) p(x) = 5x – π , x = ^{4}/_{5}

No, 5x – π = 5(^{4}/_{5}) – π= 4 – π ≠ 0 at x = ^{4}/_{5}

(iii) p(x) = x^{2} – 1 , x = 1, – 1

Yes. x^{2} – 1 = 1^{2} – 1 = 0 at x = 1,

x^{2} – 1 = (-1)^{2} – 1 = 1 – 1 = 0 at x = – 1

(iv) p(x) = (x + 1)(x – 2), x = -1, 2

yes.

(x + 1)(x – 2) = (-1 + 1)(-1 – 1) at x = -1,

(x + 1)(x – 2) = (2 + 1)(2 – 2) = 3×0 = 0 at x = 2

(v) p(x) = x^{2} , x = 0

Yes. x^{2} = 0^{2} = 0

(vi) p(x) = lx + m , x = –^{m}/_{l}

Yes. l(-^{m}/_{l}) + m = -m + m = 0 at x = –^{m}/_{l}

(vii) p(x) = 3x^{2} – 1 , x = – ^{1}/_{√3}, ^{2}/_{√3}

3x^{2} – 1 = 3(- ^{1}/_{√3})^{2} – 1 = 3(^{1}/_{3}) – 1 = 0 at x = – ^{1}/_{√3}

3x^{2} – 1 = 3(- ^{2}/_{√3})^{2} – 1 = 3(^{4}/_{3}) – 1 = 4 – 1 = 3≠0 at –^{2}/_{√3}

Therefore, at – ^{1}/_{√3} is a zero but –^{2}/_{√3} is not a zero of the polynomial.

(viii) p(x) = 2x + 1, x = ^{1}/_{2}

2x + 1 = 2(^{1}/_{2}) + 1 = 2≠0 at x = ^{1}/_{2}

**Find the zero of the polynomial in each of the following cases:**

**(i) p(x) = x + 5**

**(ii) p(x) = x – 5**

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

**(iv) p(x) = 3x – 2**

**(v) p(x) = 3x**

**(vi) p(x) = ax , a ≠ 0**

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

Solution:

(i) p(x) = x + 5

x + 5 = 0 so, x = -5

Therefore, -5 is the zero of x + 5

(ii) p(x) = x – 5

x – 5 = 0 so, x = 5

Therefore, 5 is the zero of x – 5.

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

2x + 5 = 0 so, x = –^{5}/_{2}

Therefore, –^{5}/_{2} is the zero of 2x + 5

(iv) p(x) = 3x – 2

3x – 2 = 0 so, x = ^{2}/_{3}

Therefore, ^{2}/_{3} is the zero of 3x – 2

(v) p(x) = 3x

3x = 0 so, x = 0

Therefore, 0 is the zero of 3x

(vi) p(x) = ax , a ≠ 0

ax = 0 so, a = ^{0}/_{x}

Therefore, ^{0}/_{x} is the zero of ax

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

cx + d = 0 so, x = –^{d}/_{c}

Therefore, –^{d}/_{c} is the zero of cx + d