
1. Find the value of the polynomial 5x-4x2+3 at
(i) x=0
(ii) x=-1
(iii) x=2
solution :
(i) x=0
Let,
p(x)=5x-4x2+3
The value of the polynomial p(x) at x=0 is given
p(0)=5×(0)-4×(0)2+3
=0-0+3
=3
(ii) x=-1
Let,
p(x)=5x-4x2+3
The value of the polynomial p(x) at x=-1 is given
p(-1)=5×(-1)-4×(-1)2+3
=(-5)-4+3
=-5-4+3
=-9+3
=-6
(iii) x=2
Let,
p(x)=5x-4x2+3
The value of the polynomial p(x) at x=2 is given
p(2)=5×(2)-4×(2)2+3
=10-(4×4)+3
=10-16+3
=-6+3
=-3
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)
solution :
(i) p(y)=y2-y=1
p(y)=y2-y=1
When, p(0)
p(0)=02-0+1
=0-0+1
=1
When, p(1)
p(1)=12-1+1
= 1 − 1 + 1
=0+1
=1
When, p(2)
p(2)=22-2+1
=4-2+1
2+1
=3
(ii) p(t)=2+t+2t2−t3
p(t)=2+t+2t2−t3When, p(0)
p(0)=2+0+2×(0)2−03
=2+0−0
=2
When, p(1)
p(1)=2+1+2×(1)2−13
=3+2-1
=5-1
=4
When, p(2)
p(2)=2+2+2×(2)2-23
=4+8-8
=4+0
=4
(iii) p(x)=x3
p(x)=x3When, p(0)
P(0)=03
=0
When, p(1)
p(1)=13
=1
When, P(2)
p(2)=23
=8
(iv) P(x)=(x−1)(x+1)
P ( x ) = ( x − 1 ) ( x + 1 )
When, p(0)
p(0)=(0-1)(0+1)
=-1×1
=−1
When, p(1)
p(1)=(1-1)(1+1)
=0×2
=0
When, p(2)
p(2)=(2-1)(2+1)
=1×3
=3
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1,x=-13
(ii) p(x)=5x-Ï€,x=45
(iii) 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=-ml
(vii) p(x)=3x2-1, x=-1√3,2√3
(viii) p(x)=2x+1,x=12
solution :
(i) p(x)=3x+1,x=-13
For, x=-13,
p(x)=3x+1
p(-13)=3(-13)+1
=−1+1
=0
So, -13 is a zero of p(x).
(ii) p(x)=5x-Ï€,x=45
For, x=45,
p(x)=5x-Ï€
p(45)=5×(45)-π
=4−π
So, 45 is not a zero of p(x).
(iii) p(x)=x2-1,x=1,-1
For, x=1,
p(x)=x2-1,
p(1)=12-1
=1−1
=0
For, x=-1,
p(x)=x2-1,
p(-1)=-12-1
=1−1
=0
So, 1 and -1 is a zero of p(x).
(iv) p(x)=(x+1)(x-2), x=-1,2
For, x=-1,
p(x)=(x+1)(x-2),
p(-1)=(-1+1)(-1–2)
=0×(-3)
=0
For, x=2,
p(x)=(x+1)(x-2),
p(2)=(2+1)(2–2)
=3×0
=0
So, -1 and 2 is a zero of p(x).
(v) p(x)=x2,x=0
For, x=0,
p(x)=x2
p(0)=02
=0
So, 0 is a zero of p(x).
(vi) p(x)=lx+m,x=-ml
For, x=-ml
p(x)=lx+m
p(-ml)=l(-ml)+m
=-m+m
=0
So, -ml is a zero of p(x).
(vii) p(x)=3x2-1 , x=-1√3,2√3
For, x=-1√3
p(x)=3x2-1
p(-1√3)=3(-1√3)2-1
=3×(13)-1
=1-1
=0
For, x=2√3
p(x)=3x2-1
p(2√3)=3(2√3)2-1
=3×(43)-1
=4-1
=3
=3≠0
So, -1√3 is a zero of p(x) but (2√3) is not a zero of p(x).
(viii) p(x)=2x+1,x=12
For, x=12,
p(x)=2x+1
p(12)=2(12)+1
=1+1
=2
=2≠0
So, 12 is not a zero of p(x).
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
⇒x+5=0
⇒x=−5
So, -5 is a zero of the polynomial p(x).
(ii) p(x)=x–5
p(x)=x–5
⇒x-5=0
⇒x=5
So, 5 is a zero of the polynomial p(x).
(iii) p(x)=2x+5
⇒2x+5=0
⇒2x=-5
⇒x=-52
So, -52 is a zero of the polynomial p(x).
(iv) p(x)=3x-2
⇒3x-2=0
⇒3x=2
⇒x=23
So, 23 is a zero of the polynomial p(x).
(v) p(x)=3x
⇒3x=0
⇒x=03
⇒x=0
So, 0 is a zero of the polynomial p(x).
(vi) p(x)=ax, a≠0
⇒ax=0
⇒x=0a
⇒x=0
So, 0 is a zero of the polynomial p(x).
(vii) p(x)=cx+d, c≠0,c,d are real numbers.
⇒cx+d=0
⇒cx=−d
⇒x=-dc
So, -dc is a zero of the polynomial p(x).
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