Q1. Find the value of the polynomial 5x – 4x^{2} + 3 at
(i) x = 0
(ii) x = –1
(iii) x = 2
Ans:
Let f(x) = 5x−4x^{2}+3
(i) When x = 0
f(0) = 5(0)4(0)^{2}+3
= 3
(ii) When x = 1
f(x) = 5x−4x^{2}+3
f(−1) = 5(−1)−4(−1)^{2}+3
= −5–4+3
= −6
(iii) When x = 2
f(x) = 5x−4x^{2}+3
f(2) = 5(2)−4(2)^{2}+3
= 10–16+3
= −3
Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y^{2 }− y + 1
Ans: 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 = 3
(ii) p(t) = 2 + t + 2t^{2 }− t^{3}
Ans: 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}
Ans: ∴ p(0) = (0)^{3} = 0
p(1) = (1)^{3} = 1
p(2) = (2)^{3} = 8
(iv) P(x) = (x − 1) (x + 1)
Ans: ∴ p(0) = (0 – 1)(0 + 1) = (−1)(1) = –1
p(1) = (1 – 1)(1 + 1) = 0(2) = 0
p(2) = (2 – 1)(2 + 1) = 1(3) = 3
Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = −1/3
Ans: For, x = 1/3, p(x) = 3x+1
∴ p(−1/3) = 3(1/3)+1 = −1+1 = 0
∴ 1/3 is a zero of p(x).
(ii) p(x) = 5x – π, x = 4/5
Ans: For, x = 4/5, p(x) = 5x – π
∴ p(4/5) = 5(4/5)  π = 4 π
∴ 4/5 is not a zero of p(x).
(iii) p(x) = x^{2 }− 1, x = 1, −1
Ans: For, x = 1, −1;
p(x) = x^{2}−1
∴ p(1) = 1^{2 }− 1 = 1 − 1 = 0
p(−1) = (1)^{2 }− 1 = 1 − 1 = 0
∴ 1, −1 are zeros of p(x).
(iv) p(x) = (x+1)(x–2), x =−1, 2
Ans: For, x = −1,2;
p(x) = (x+1)(x–2)
∴ p(−1) = (−1+1)(−1–2)
= (0)(−3) = 0
p(2) = (2+1)(2–2) = (3)(0) = 0
∴ −1,2 are zeros of p(x).
(v) p(x) = x^{2}, x = 0
Ans: For, x = 0 p(x) = x^{2}
p(0) = 0^{2} = 0
∴ 0 is a zero of p(x).
(vi) p(x) = lx + m, x = −m/l
Ans: For, x = m/l ; p(x) = lx+m
∴ p(m/l)= l(m/l)+m = −m+m = 0
∴ m/l is a zero of p(x).
(vii) p(x) = 3x^{2}−1, x = 1/√3 , 2/√3
Ans: For, x = 1/√3 , 2/√3 ; p(x) = 3x^{2}−1
∴ p(1/√3) = 3(1/√3)^{2}1 = 3(1/3)1 = 11 = 0
∴ p(2/√3 ) = 3(2/√3)^{2}1 = 3(4/3)1 = 4−1=3 ≠ 0
∴ 1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).
(viii) p(x) =2x + 1, x = 1/2
Ans: For, x = 1/2 p(x) = 2x + 1
∴ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0
∴ 1/2 is not a zero of p(x).
Q4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Ans: p(x) = x + 5
⇒ x + 5 = 0
⇒ x = −5
∴ 5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = x – 5
Ans: p(x) = x − 5
⇒ x − 5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x + 5
Ans: p(x) = 2x + 5
⇒ 2x+5 = 0
⇒ 2x = −5
⇒ x = 5/2
∴ x = 5/2 is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3x–2
Ans: p(x) = 3x–2
⇒ 3x − 2 = 0
⇒ 3x = 2
⇒x = 2/3
∴ x = 2/3 is a zero polynomial of the polynomial p(x).
(v) p(x) = 3x
Ans: p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a ≠ 0
Ans: p(x) = ax
⇒ ax = 0
⇒ x = 0
∴ x = 0 is a zero polynomial of the polynomial p(x).
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Ans: p(x) = cx + d
⇒ cx + d =0
⇒ x = d/c
∴ x = d/c is a zero polynomial of the polynomial p(x).
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