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Question

Find the zero of the polynomial in each of the following cases:
(i) p(x)=x+5
(ii) p(x)=x5
(iii) p(x)=2x+5
(iv) p(x)=3x2
(v) p(x)=3x
(vi) p(x)=ax,a0
(vii) p(x)=cx+d,c0 and c,d are real numbers.

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Solution

To get the zero of polynomial p(x), take p(x)=0 and hence solve the equation.

(i) p(x)=x+5

x+5=0

x=5

5 is the zero of p(x)=x+5.

(ii) p(x)=x5

x+5=0

x=5

5 is the zero of p(x)=x5.

(iii) p(x)=2x+5

2x+5=0

2x=5

x=52

52 is the zero of p(x)=2x+5

(iv) p(x)=3x2

3x2=0

3x=2

x=23

So 23 is the zero of p(x)=3x2.

(v) p(x)=3x

3x=0

x=0

So 0 is the zero of p(x)=3x.

(vi) p(x)=ax,a0

ax=0 ,a0

x=0a=0

So, 0 is the zero of .p(x)=ax,a0

(vii) p(x)=cx+d,c0 and c,d are real numbers.

cx+d=0,(c0)

cx=d

x=dc

So, dc is the zero of p(x)=cx+d.


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