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Answers
Answer:
The zero of the polynomial is defined as any real value of x, for which the value of the polynomial becomes zero. A real number a will be a zero of p(x) if p(a) =0
___________
i) To check whether 4 is a zero of p(x)= x-4 , we will write 4 in place of x.
p(4)=4-4
=0
As p(4)=0 , 4 is a zero of the polynomial.
Hence verified
___________
ii) We will use same process as above to verify if -3 is a zero of p(x).
p(-3)=-3+3
=0
As p(-3)=0 , -3 is a zero of polynomial p(x).
___________
iii) Similarly,
We will put -1/2 in p(x)
Hence, -1/2 is a zero of p(x).
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iv) Now we will first put x=1 in p(x).
p(1) = (1)²-3(1) +2
= 1-3+2
= 3-3
=0
Again, we will put x = 2 in p(x).
p(2)= (2)² -3(2)+2
= 4 -6+2
= 6-6
=0
Hence, 1 and 2 are zeroes of p(x)
____________
v) We have, p(x)=x²+x-6
p(2)=(2)²+2-6
= 4+2-6
= 6-6
=0
So 2 is a zero of p(x)
Now,
p(-3)=(-3)²+(-3)-6
= 9-3-6
= 9-9
=0
-3 is zero of p(x)
Hence verified.
____________
vi) p(x) =x²-3x
p(0)= (0)²-3(0)
=0
So , 0 is a zero of p(x)
p(3)= (3)²-3(3)
= 9-9
=0
3 is also a zero of p(x)
Hence verified
_____________
vii) p(x) = x³-3x² +2x
p(0)=(0)³-3(0)+2(0)= 0
p(1)= (1)³ -3(1)+2(1)
= 1-3+2
= 3-3=0
p(2) =(2)³ -3(2)²+2(2)
= 8 -12 +4
= 12-12 =0
Since p(0), p(1) and p(2) are equal to 0 , It means that 0,1 and 2 are the zeroes of the p(x).
______________
viii) p(x) = x³+x²-9x-9
p(0) = (0)³ +(0)²-9(0)-9
= 0-9
= -9
p(3) = (3)³ +(3)²-9(3)-9
= 27+9-27-9
=0
p(-3) = (-3)³ +(-3)²-9(-3)-9
= -27+9+27-9
=0
p(-1) = (-1)³ +(-1)²-9(-1)-9
= -1+1+9-9
= 0
Since p(3), p(-3) & p(-1) are equal to 0 , Hence -1, -3 & 3 are the zeroes of the polynomial. p(0) is not equal to 0 and that is why it is not a zero of polynomial.
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