Answer :
Let's solve each of the given polynomial equations step by step to find their zeros.
### (i) [tex]\(P(x) = x - 5\)[/tex]
To find the zero of the polynomial, we need to determine the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = x - 5 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ x - 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the zero of the polynomial [tex]\(P(x) = x - 5\)[/tex] is [tex]\( x = 5 \)[/tex].
### (ii) [tex]\(P(x) = 2x + 5\)[/tex]
Similarly, we need to find the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = 2x + 5 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ 2x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 2x = -5 \][/tex]
[tex]\[ x = -\frac{5}{2} \][/tex]
So, the zero of the polynomial [tex]\(P(x) = 2x + 5\)[/tex] is [tex]\( x = -\frac{5}{2} \)[/tex].
### (iii) [tex]\(P(x) = 3\)[/tex]
Here, we need to find the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = 3 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ 3 = 0 \][/tex]
This is a contradiction because 3 can never equal 0. Therefore, there are no values of [tex]\(x\)[/tex] that satisfy this equation.
So, the polynomial [tex]\(P(x) = 3\)[/tex] has no zeros.
### (iv) [tex]\(P(x) = cx + d\)[/tex], [tex]\(c \neq 0\)[/tex]
To find the zero, we set the polynomial equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ P(x) = cx + d \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ cx + d = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ cx = -d \][/tex]
[tex]\[ x = -\frac{d}{c} \][/tex]
So, the zero of the polynomial [tex]\(P(x) = cx + d\)[/tex] with [tex]\(c \neq 0\)[/tex] is [tex]\( x = -\frac{d}{c} \)[/tex].
### Summary:
- The zero of [tex]\(P(x) = x - 5\)[/tex] is [tex]\( x = 5 \)[/tex].
- The zero of [tex]\(P(x) = 2x + 5\)[/tex] is [tex]\( x = -\frac{5}{2} \)[/tex].
- The polynomial [tex]\(P(x) = 3\)[/tex] has no zeros.
- The zero of [tex]\(P(x) = cx + d\)[/tex] with [tex]\(c \neq 0\)[/tex] is [tex]\( x = -\frac{d}{c} \)[/tex].
### (i) [tex]\(P(x) = x - 5\)[/tex]
To find the zero of the polynomial, we need to determine the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = x - 5 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ x - 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the zero of the polynomial [tex]\(P(x) = x - 5\)[/tex] is [tex]\( x = 5 \)[/tex].
### (ii) [tex]\(P(x) = 2x + 5\)[/tex]
Similarly, we need to find the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = 2x + 5 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ 2x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 2x = -5 \][/tex]
[tex]\[ x = -\frac{5}{2} \][/tex]
So, the zero of the polynomial [tex]\(P(x) = 2x + 5\)[/tex] is [tex]\( x = -\frac{5}{2} \)[/tex].
### (iii) [tex]\(P(x) = 3\)[/tex]
Here, we need to find the value of [tex]\(x\)[/tex] that makes [tex]\(P(x)\)[/tex] equal to zero.
[tex]\[ P(x) = 3 \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ 3 = 0 \][/tex]
This is a contradiction because 3 can never equal 0. Therefore, there are no values of [tex]\(x\)[/tex] that satisfy this equation.
So, the polynomial [tex]\(P(x) = 3\)[/tex] has no zeros.
### (iv) [tex]\(P(x) = cx + d\)[/tex], [tex]\(c \neq 0\)[/tex]
To find the zero, we set the polynomial equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ P(x) = cx + d \][/tex]
Setting [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ cx + d = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ cx = -d \][/tex]
[tex]\[ x = -\frac{d}{c} \][/tex]
So, the zero of the polynomial [tex]\(P(x) = cx + d\)[/tex] with [tex]\(c \neq 0\)[/tex] is [tex]\( x = -\frac{d}{c} \)[/tex].
### Summary:
- The zero of [tex]\(P(x) = x - 5\)[/tex] is [tex]\( x = 5 \)[/tex].
- The zero of [tex]\(P(x) = 2x + 5\)[/tex] is [tex]\( x = -\frac{5}{2} \)[/tex].
- The polynomial [tex]\(P(x) = 3\)[/tex] has no zeros.
- The zero of [tex]\(P(x) = cx + d\)[/tex] with [tex]\(c \neq 0\)[/tex] is [tex]\( x = -\frac{d}{c} \)[/tex].