At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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).
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
