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Q.1. Find the value of the polynomial 5x – 4x^{2} + 3 at
(i) x = 0
(ii) x = –1
(iii) x = 2
Solution:
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
Q.2. 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(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}
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)^{3} = 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)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3
Q.3. Verify whether the following are zeros of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =1/3
(ii) p(x) = 5x – p , 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
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
For, x = 4/5, p(x) = 5x–π
∴ p(4/5) = 5(4/5) = 4
∴ 4/5 is not a zero of p(x).
(iii) p(x)=x^{2}−1, x=1, −1
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
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
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
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
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
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).
Zero Polynomial
Q.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.
Solution:
(i) p(x) = x+5
p(x) = x+5
⇒ x+5 = 0
⇒ x = −5
∴ 5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = x–5
p(x) = x−5
⇒ x−5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x+5
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
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
p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a0
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.
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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