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ZEROES OF A POLYNOMIALLet p(x) is a polynomial. If p(a) = 0, then â€˜aâ€™ is said to be a zero of the polynomial p(x).

Note: (i) Finding the â€˜zerosâ€™ of a polynomial means solving the equation p(x) = 0.

(ii) A non-zero â€˜constant polynomialâ€™ has no zero.

(iii) Every real number is a zero of the zero polynomial.

(iv) Every linear polynomial has one and only one zero.

(v) A polynomial can have more than one zero.

(vi) A zero of a polynomial need not be 0.

(vii) â€˜0â€™ may be a zero of a polynomial.

**Ques 1. Find the value of the polynomial 5x â€“ 4x ^{2} + 3 at (i) x = 0 (ii) x = â€“1 (iii) x = 2**

**Ans: **(i) âˆµ p(x) = 5x â€“ 4x^{2} + 3 = 5(x) â€“ 4(x)^{2} + 3

âˆ´ p(0) = 5(0) â€“ 4(0) + 3 = 0 â€“ 0 + 3 = 3

Thus, the value of 5x â€“ 4x^{2} + 3 at x = 0 is 3.

(ii) âˆµ p(x) = 5x â€“ 4x^{2} + 3 = 5(x) â€“ 4(x)^{2} + 3

âˆ´ p(â€“1) = 5(â€“1) â€“ 4(â€“1)^{2 }+ 3 = â€“5 â€“ 4(1) + 3 = â€“5 â€“ 4 + 3 = â€“9 + 3 = â€“6

âˆ´ The value of 5x â€“ 4x^{2} + 3 at x = â€“1 is â€“6.

(iii) âˆµ p(x) = 5x â€“ 4x^{2} + 3 = 5(x) â€“ 4(x)^{2} + 3

âˆ´ p(2) = 5(2) â€“ 4(2)^{2} + 3 = 10 â€“ 4(4) + 3 = 10 â€“ 16 + 3 = â€“3

Thus the value of 5x â€“ 4x^{2} + 3 at x = 2 is â€“3.

**Ques 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y ^{2} â€“ y + 1 (ii) p(t) = 2 + t + 2t^{2} â€“ t^{3} (iii) p(x) = x^{3} (iv) p(x) = (x â€“ 1) (x + 1)**

**Ans:** (i) p(y) = y^{2} â€“ y + 1

âˆµ p(y) = y^{2} â€“ y + 1 = (y)^{2 }â€“ y + 1

âˆ´ p(0) = (0)^{2} â€“ (0) + 1 = 0 â€“ 0 + 1 = 1

p(1) = (1)^{2 }â€“ (1) + 1 = 1 â€“ 1 + 1 = 1

p(2) = (2)^{2} â€“2 + 1 = 4 â€“ 2 + 1 = 3

(ii) p(t) = 2 + t + 2t2 â€“ t3

âˆµ p(t) = 2 + t + 2t^{2} â€“ t^{3} = 2 + t + 2(t)^{2} â€“ (t)^{3}

âˆ´ p(0) = 2 + (0) + 2(0)^{2} â€“ (0)^{3} = 2 + 0 + 0 â€“ 0 = 2

p(1) = 2 + (1) + 2 (1)^{2 }â€“ (1)^{3 }= 2 + 1 + 2 â€“ 1 = 4

p(2) = 2 + 2 + 2(2)^{2} â€“ (2)^{3} = 2 + 2 + 8 â€“ 8 = 4

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

âˆ´ p(0) = (0)^{3 }= 0 p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8 [âˆµ 2 x^{2} x^{2} = 8]

(iv) p(x) = (x â€“ 1)(x + 1)

âˆµ p(x) = (x â€“ 1)(x + 1)

âˆ´ p(0) = (0 â€“ 1)(0 + 1) = â€“1 x 1 = â€“1

p(1) = (1 â€“ 1)(1 + 1) = (0)(2) = 0

p(2) = (2 â€“ 1)(2 + 1) = (1)(3) = 3

**Ques 3. Verify whether the following are zeros of the polynomial, indicated against them. **

(i) p(x) = 3x + 1, x =

(ii) p(x) = 5x â€“ p , x =

(iii) p(x) = x^{2} â€“ 1, x = 1, â€“ 1

(iv) p(x) = (x + 1) (x â€“ 2), x = â€“1, 2

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

(vi) p(x) = lx + m, x =

(vii) p(x) = 3x^{2} â€“ 1, x =

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

**Ans:** (i) âˆµ p(x) = 3x + 1

(iii) Since, p(x) = x^{2} â€“ 1

âˆ´ p(1) = (1)^{2} â€“ 1 = 1 â€“ 1 = 0

Since, p(1) = 0,

âˆ´ x = 1 is a zero of x^{2} â€“ 1.

Also p(â€“1) = (â€“1)^{2 }â€“ 1 = 1 â€“ 1 = 0 i.e. p(â€“1) = 0,

âˆ´ x = â€“1 is also a zero of x^{2} â€“ 1.

(iv) We have p(x) = (x + 1)(x â€“ 2)

âˆ´ p(â€“1) = (â€“1 + 1)(â€“1 â€“ 2) = (0)(â€“3) = 0

Since p(â€“1) = 0,

âˆ´ x = â€“1 is a zero of (x + 1) (x â€“ 1).

Also, p(2) = (2 + 1)(2 â€“ 2) = (3)(0) = 0

Since p(2) = 0,

âˆ´ x = 2 is also a zero of (x + 1)(x â€“ 1).

(v) We have p(x) = x^{2 }

âˆ´ p(0) = (0)^{2} = 0 Since p(0) = 0,

âˆ´ 0 is a zero of x^{2}.

**Ques 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x â€“ 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x â€“ 2 ( v ) p(x) = 3x (vi) p(x) = ax, a â‰ 0 (vii) p(x) = cx + d, c â‰ 0, c, d are real numbers. Ans: **(i) We have p(x) = x + 5

âˆ´ p(x) = 0

â‡’ x + 5 = 0 or x = â€“5

Thus, a zero of x + 5 is (â€“5).

(ii) We have p(x) = x â€“ 5

âˆ´ p(x) = 0

â‡’ x â€“ 5 = 0 or x = 5

Thus, a zero of x â€“ 5 is 5.

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

âˆ´ p(x) = 0

â‡’ 2x + 5 = 0 or 2x = â€“5 or x =

Thus, a zero of 2x + 5 is

(iv) Since p(x) = 3x â€“ 2 âˆ´ p(x) = 0

â‡’ 3x â€“ 2 = 0 or 3x = 2 or x = 2/3

Thus, a zero of 3x â€“ 2 is 2/3

(v) Since p(x) = 3x âˆ´ p(x) = 0

â‡’ 3x = 0 or x = 0/3 or 0

Thus, a zero of 3x is 0.

(vi) Since, p(x) = ax, a **â‰ ** 0

â‡’ p(x) = 0

â‡’ ax = 0 or x = 0 a = 0

Thus, a zero of ax is 0.

(vii) Since, p(x) = cx + d

âˆ´ p(x) = 0

â‡’ cx + d = 0 or c x = â€“d or x = â€“ dc

Thus, a zero of cx + d is â€“ dc.

**Ques 5: If p(x) = x2 â€“ 4x + 3, evaluate: p(2) â€“ p(â€“1) + **

**Ans: **We have p(x) = x2 â€“ 4x + 3

âˆ´ p(â€“1) = (â€“1)^{2} â€“ 4(â€“1) + 3

= 1 + 4 + 3 = 8

and p(2) = (2)^{2} â€“ 4(2) + 3 = 4 â€“ 8 + 3 = â€“1

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