TOPICS

Exercise - 2.2

Polynomials

**Question-1 :-** Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0,

(ii) x = –1,

(iii) x = 2.

(i) At x = 0 p(x) = 5x – 4x² + 3 p(0) = 5 x 0 - 4 x 0 + 3 = 0 - 0 + 3 = 3

(ii) At x = -1 p(x) = 5x – 4x² + 3 p(-1) = 5 x (-1) – 4 x (-1)² + 3 = -5 - 4 x 1 + 3 = -5 - 4 + 3 = -9 + 3 = -6

(iii) At x = 2 p(x) = 5x – 4x² + 3 p(2) = 5 x 2 - 4 x 2² + 3 = 10 - 4 x 4 + 3 = 13 - 16 = -3

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

(i) p(y) = y² – y + 1,

(ii) p(t) = 2 + t + 2t² – t³,

(iii) p(x) = x³,

(iv) p(x) = (x – 1) (x + 1).

(i) p(y) = y² – y + 1 at p(0) = 0² – 0 + 1 = 0 - 0 + 1 = 1 p(1) = 1² – 1 + 1 = 1 - 1 + 1 = 2 - 1 = 1 p(2) = 2² – 2 + 1 = 4 - 2 + 1 = 5 - 2 = 3

(ii) p(t) = 2 + t + 2t² – t³ p(0) = 2 + 0 + 2 x 0² – 0³ = 2 + 0 = 2 p(1) = 2 + 1 + 2 x 1² – 1³ = 3 + 2 - 1 = 5 - 1 = 4 p(2) = 2 + 2 + 2 x 2² – 2³ = 4 + 2 x 4 - 8 = 4 + 8 - 8 = 12 - 8 = 4

(iii) p(x) = x³ p(0) = 0³ = 0 p(1) = 1³ = 1 p(2) = 2³ = 8

(iv) p(x) = (x – 1) (x + 1) p(0) = (0 – 1) (0 + 1) = -1 x 1 = -1 p(1) = (1 – 1) (1 + 1) = 0 x 2 = 0 p(2) = (2 – 1) (2 + 1) = 1 x 3 = 3

**Question-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) = p(x) = x² – 1, x = 1, –1

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

(v) p(x) = x², x = 0

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

(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3

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

(i) p(x) = 3x + 1, x = -1/3 p(-1/3) = 3 x (-1/3) + 1 = -1 + 1 = 0 Hence, x = -1/3 is a zero of polynomial.

(ii) p(x) = 5x - π, x = 4/5 p(4/5) = 5 x 4/5 - π = 4 - π Hence, x = 4/5 is not a zero of polynomial.

(iii) p(x) = x² – 1, x = 1, –1 p(1) = (1)² – 1 = 1 - 1 = 0 p(-1) = (-1)² – 1 = 1 - 1 = 0 Hence, x = 1, -1 are zeroes of polynomial.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2 p(-1) = (-1 + 1)(-1 - 2) = 0 x (-3) = 0 p(2) = (2 + 1)(2 - 2) = 3 x 0 = 0 Hence, x = -1, 2 are zeroes of polynomial.

(v) p(x) = x², x = 0 p(0) = 0² = 0 Hence, x = 0 is a zero of polynomial.

(vi) p(x) = lx + m, x = -m/l p(-m/l) = l x (-m/l) + m = -m + m = 0 Hence, x = -m/l is a zero of polynomial.

(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3 p(-1/√3) = 3 x (-1/√3)² – 1 = 3 x 1/3 - 1 = 1 - 1 = 0 p(2/√3) = 3 x (2/√3)² – 1 = 3 x 4/3 - 1 = 4 - 1 = 3 Hence, x = -1/√3 is a zero of polynomial. But x = 2/√3 is not a zero of polynomial.

(viii) p(x) = 2x + 1, x = 1/2 p(1/2) = 2 x 1/2 + 1 = 1 + 1 = 2 Hence, x = 2 is not a zero of polynomial.

**Question-4 :-** 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.

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0. So, p(x) = 0. (i) p(x) = x + 5 p(x) = 0 x + 5 = 0 x = -5 Therefore, x = -5 is a zero of polynomial p(x) = x + 5.

(ii) p(x) = x – 5 p(x) = 0 x - 5 = 0 x = 5 Therefore, x = 5 is a zero of polynomial p(x) = x - 5.

(iii) p(x) = 2x + 5 p(x) = 0 2x + 5 = 0 2x = -5 x = -5/2 Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5.

(iv) p(x) = 3x – 2 p(x) = 0 3x - 2 = 0 3x = 2 x = 2/3 Therefore, x = 2/3 is a zero of polynomial p(x) = 3x - 2.

(v) p(x) = 3x p(x) = 0 3x = 0 x = 0 Therefore, x = 0 is a zero of polynomial p(x) = 3x.

(vi) p(x) = ax, a ≠ 0 p(x) = 0 ax = 0 x = 0/a x = 0 Therefore, x = 0 is a zero of polynomial p(x) = ax.

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. p(x) = 0 cx + d = 0 cx = -d x = -d/c Therefore, x = -d/c is a zero of polynomial p(x) = cx + d.

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