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
Answers
Answer:
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Step-by-step explanation:
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Answer:
i)-5
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¡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