Question

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​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

i)-5

¡¡)5

¡¡¡)-5/2

¡v)2/3

v)0

vi)0

vii)-d/c

Step-by-step explanation:

Solution: (i)p(x) = x + 5

Plug p(x) = 0 we get

=>x+5 =0

=> x = - 5

-5 is zero of the polynomial

(ii) p(x) = x – 5

Plug p(x) = 0 we get

=>x - 5 = 0

=> x = 5

5 is zero of the polynomial

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

Plug p(x) = 0 we get

=>2x+5 =0

=> 2x = - 5

=> x= -5/2

-5/2is zero of the polynomial

(iv) p(x) = 3x – 2

Plug p(x) = 0 we get

=>3x-2=0

=>3x =2

=> x = 2/3

2/3 is zero of the polynomial

(v) p(x) = 3x

Plug p(x) = 0 we get

=>3x=0

=> x= 0/3

=> x=0

0 is zero of the polynomial

(vi) p(x) = ax, a ≠ 0

Plug p(x) = 0 we get

=>ax=0

=> x =0/a

=> x = 0

0 is zero of the polynomial

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Plug p(x) = 0 we get

=>cx +d = 0

=> cx = - d

=> x = -d/c

-d/c is zero of the polynomial