Polynomials
Chapter 2
Exercise 2.2
EX 2.2 QUESTION 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p(x) = 5x – 4x2 + 3
(i) p(0)
= 5(0) – 4(0)2 + 3
= 0 – 0 + 3 = 3
(ii) p(-1)
= 5(-1) – 4(-1)2 + 3
= – 5x – 4x2 + 3
= -9 + 3 = -6
(iii) p(2)
= 5(2) – 4(2)2 + 3
= 10 – 4(4) + 3
= 10 – 16 + 3 = -3
EX 2.2 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 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:
(i) p(y) = y2–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+2t2−t3:
p(t) = 2+t+2t2−t3
∴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)=x3
p(x) = x3
∴p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) P(x) = (x−1)(x+1)
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
EX 2.2 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) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = -1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = -m/l
(vii) P (x) = 3x2 – 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) –π = 4 –π
∴ 4/5 is not a zero of p(x).
(iii) p(x)=x2−1,x=1, −1
For, x = 1, −1
p(x) = x2−1
∴p(1)=12−1=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) = x2, x = 0
For, x = 0 p(x) = x2
p(0) = 02 = 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) = 3x2−1, x = -1/√3 , 2/√3
For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 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).
EX 2.2 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 where c and d are real numbers.
Solution
(i) p(x) = x+5
⇒ x+5 = 0
⇒ x = −5
∴ -5 is a zero of the polynomial p(x).
(ii)p(x) = x−5
⇒ x−5 = 0
⇒ x = 5
∴ 5 is a zero 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 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 of the polynomial p(x).
(v) p(x) = 3x
p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴0 is a zero of the polynomial p(x).
(vi) p(x) = ax, a0
p(x) = ax
⇒ ax = 0
⇒ x = 0
∴x = 0 is a zero 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 of the polynomial p(x).