Answer :
Certainly! Let's verify whether the given values of [tex]\(x\)[/tex] are indeed the zeroes of the given polynomials step-by-step.
### (i) [tex]\(p(x) = 2x + 1\)[/tex]; [tex]\(x = -\frac{1}{2}\)[/tex]
We substitute [tex]\(x = -\frac{1}{2}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) + 1 = -1 + 1 = 0 \][/tex]
So, [tex]\( x = -\frac{1}{2} \)[/tex] is a zero of the polynomial.
### (ii) [tex]\(p(x) = 5x - \pi\)[/tex]; [tex]\(x = -\frac{3}{2}\)[/tex]
We substitute [tex]\(x = -\frac{3}{2}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{3}{2}\right) = 5\left(-\frac{3}{2}\right) - \pi = -\frac{15}{2} - \pi \neq 0 \][/tex]
So, [tex]\( x = -\frac{3}{2} \)[/tex] is not a zero of the polynomial.
### (iii) [tex]\(p(x) = x^2 - 1\)[/tex]; [tex]\(x = \pm 1\)[/tex]
For [tex]\(x = 1\)[/tex]:
[tex]\[ p(1) = 1^2 - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ p(-1) = (-1)^2 - 1 = 1 - 1 = 0 \][/tex]
So, both [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex] are zeroes of the polynomial.
### (iv) [tex]\(p(x) = (x-1)(x+2)\)[/tex]; [tex]\(x = -1, -2\)[/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ p(-1) = (-1 - 1)(-1 + 2) = (-2)(1) = -2 \neq 0 \][/tex]
For [tex]\(x = -2\)[/tex]:
[tex]\[ p(-2) = (-2 - 1)(-2 + 2) = (-3)(0) = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a zero, but [tex]\( x = -2 \)[/tex] is a zero of the polynomial.
### (v) [tex]\(p(y) = y^2\)[/tex]; [tex]\(y = 0\)[/tex]
We substitute [tex]\(y = 0\)[/tex] into [tex]\(p(y)\)[/tex]:
[tex]\[ p(0) = 0^2 = 0 \][/tex]
So, [tex]\( y = 0 \)[/tex] is a zero of the polynomial.
### (vi) [tex]\(p(x) = ax + b\)[/tex]; [tex]\(x = -\frac{b}{a}\)[/tex]
We substitute [tex]\(x = -\frac{b}{a}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{b}{a}\right) = a\left(-\frac{b}{a}\right) + b = -b + b = 0 \][/tex]
So, [tex]\( x = -\frac{b}{a} \)[/tex] is a zero of the polynomial, regardless of the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
### (vii) [tex]\(f(x) = 3x^2 - 1\)[/tex]; [tex]\(x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)[/tex]
For [tex]\(x = -\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ f\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1 = 3 \left(\frac{1}{3}\right) - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = \frac{2}{\sqrt{3}}\)[/tex]:
[tex]\[ f\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1 = 3 \left(\frac{4}{3}\right) - 1 = 4 - 1 = 3 \neq 0 \][/tex]
So, [tex]\( x =- \frac{1}{\sqrt{3}} \)[/tex] is a zero, but [tex]\( x = \frac{2}{\sqrt{3}} \)[/tex] is not a zero of the polynomial.
### (viii) [tex]\(f(x) = 2x - 1\)[/tex]; [tex]\(x = \frac{1}{2}, -\frac{1}{2}\)[/tex]
For [tex]\(x = \frac{1}{2}\)[/tex]:
[tex]\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) - 1 = -1 - 1 = -2 \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{2} \)[/tex] is a zero, but [tex]\( x = -\frac{1}{2} \)[/tex] is not a zero of the polynomial.
To summarize:
1. [tex]\( x = -\frac{1}{2} \)[/tex] is a zero for (i).
2. [tex]\( x = -\frac{3}{2} \)[/tex] is not a zero for (ii).
3. [tex]\( x = \pm 1 \)[/tex] are zeros for (iii).
4. [tex]\( x = -1 \)[/tex] is not a zero, but [tex]\( x = -2 \)[/tex] is a zero for (iv).
5. [tex]\( y = 0 \)[/tex] is a zero for (v).
6. [tex]\( x = -\frac{b}{a} \)[/tex] is a zero for (vi).
7. [tex]\( x = -\frac{1}{\sqrt{3}} \)[/tex] is a zero, but [tex]\( x = \frac{2}{\sqrt{3}} \)[/tex] is not a zero for (vii).
8. [tex]\( x = \frac{1}{2} \)[/tex] is a zero, but [tex]\( x = -\frac{1}{2} \)[/tex] is not a zero for (viii).
### (i) [tex]\(p(x) = 2x + 1\)[/tex]; [tex]\(x = -\frac{1}{2}\)[/tex]
We substitute [tex]\(x = -\frac{1}{2}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) + 1 = -1 + 1 = 0 \][/tex]
So, [tex]\( x = -\frac{1}{2} \)[/tex] is a zero of the polynomial.
### (ii) [tex]\(p(x) = 5x - \pi\)[/tex]; [tex]\(x = -\frac{3}{2}\)[/tex]
We substitute [tex]\(x = -\frac{3}{2}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{3}{2}\right) = 5\left(-\frac{3}{2}\right) - \pi = -\frac{15}{2} - \pi \neq 0 \][/tex]
So, [tex]\( x = -\frac{3}{2} \)[/tex] is not a zero of the polynomial.
### (iii) [tex]\(p(x) = x^2 - 1\)[/tex]; [tex]\(x = \pm 1\)[/tex]
For [tex]\(x = 1\)[/tex]:
[tex]\[ p(1) = 1^2 - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ p(-1) = (-1)^2 - 1 = 1 - 1 = 0 \][/tex]
So, both [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex] are zeroes of the polynomial.
### (iv) [tex]\(p(x) = (x-1)(x+2)\)[/tex]; [tex]\(x = -1, -2\)[/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ p(-1) = (-1 - 1)(-1 + 2) = (-2)(1) = -2 \neq 0 \][/tex]
For [tex]\(x = -2\)[/tex]:
[tex]\[ p(-2) = (-2 - 1)(-2 + 2) = (-3)(0) = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a zero, but [tex]\( x = -2 \)[/tex] is a zero of the polynomial.
### (v) [tex]\(p(y) = y^2\)[/tex]; [tex]\(y = 0\)[/tex]
We substitute [tex]\(y = 0\)[/tex] into [tex]\(p(y)\)[/tex]:
[tex]\[ p(0) = 0^2 = 0 \][/tex]
So, [tex]\( y = 0 \)[/tex] is a zero of the polynomial.
### (vi) [tex]\(p(x) = ax + b\)[/tex]; [tex]\(x = -\frac{b}{a}\)[/tex]
We substitute [tex]\(x = -\frac{b}{a}\)[/tex] into [tex]\(p(x)\)[/tex]:
[tex]\[ p\left(-\frac{b}{a}\right) = a\left(-\frac{b}{a}\right) + b = -b + b = 0 \][/tex]
So, [tex]\( x = -\frac{b}{a} \)[/tex] is a zero of the polynomial, regardless of the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
### (vii) [tex]\(f(x) = 3x^2 - 1\)[/tex]; [tex]\(x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)[/tex]
For [tex]\(x = -\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ f\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1 = 3 \left(\frac{1}{3}\right) - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = \frac{2}{\sqrt{3}}\)[/tex]:
[tex]\[ f\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1 = 3 \left(\frac{4}{3}\right) - 1 = 4 - 1 = 3 \neq 0 \][/tex]
So, [tex]\( x =- \frac{1}{\sqrt{3}} \)[/tex] is a zero, but [tex]\( x = \frac{2}{\sqrt{3}} \)[/tex] is not a zero of the polynomial.
### (viii) [tex]\(f(x) = 2x - 1\)[/tex]; [tex]\(x = \frac{1}{2}, -\frac{1}{2}\)[/tex]
For [tex]\(x = \frac{1}{2}\)[/tex]:
[tex]\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0 \][/tex]
For [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) - 1 = -1 - 1 = -2 \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{2} \)[/tex] is a zero, but [tex]\( x = -\frac{1}{2} \)[/tex] is not a zero of the polynomial.
To summarize:
1. [tex]\( x = -\frac{1}{2} \)[/tex] is a zero for (i).
2. [tex]\( x = -\frac{3}{2} \)[/tex] is not a zero for (ii).
3. [tex]\( x = \pm 1 \)[/tex] are zeros for (iii).
4. [tex]\( x = -1 \)[/tex] is not a zero, but [tex]\( x = -2 \)[/tex] is a zero for (iv).
5. [tex]\( y = 0 \)[/tex] is a zero for (v).
6. [tex]\( x = -\frac{b}{a} \)[/tex] is a zero for (vi).
7. [tex]\( x = -\frac{1}{\sqrt{3}} \)[/tex] is a zero, but [tex]\( x = \frac{2}{\sqrt{3}} \)[/tex] is not a zero for (vii).
8. [tex]\( x = \frac{1}{2} \)[/tex] is a zero, but [tex]\( x = -\frac{1}{2} \)[/tex] is not a zero for (viii).
