Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Certainly! Let's determine the factors of the polynomial [tex]\( -3 x^3 + 15 x^2 + 3 x - 15 \)[/tex]. We'll do this by using polynomial factorization methods, specifically by testing the given factors to see if they divide the polynomial without leaving a remainder.
First, write the polynomial in question:
[tex]\[ P(x) = -3 x^3 + 15 x^2 + 3 x - 15 \][/tex]
Now let's test each of the given possible factors to see which ones are actual factors of the polynomial.
1. Factor: [tex]\( \pi + 1 \)[/tex]
- This is a constant and not a suitable factor for a polynomial in [tex]\( x \)[/tex]. Factors of a polynomial should also be polynomial expressions in [tex]\( x \)[/tex].
2. Factor: [tex]\( 3x - 1 \)[/tex]
- To test if [tex]\( 3x - 1 \)[/tex] is a factor, substitute [tex]\( x = \frac{1}{3} \)[/tex] into the polynomial:
[tex]\[ P\left(\frac{1}{3}\right) = -3 \left(\frac{1}{3}\right)^3 + 15 \left(\frac{1}{3}\right)^2 + 3 \left(\frac{1}{3}\right) - 15 = -3 \left(\frac{1}{27}\right) + 15 \left(\frac{1}{9}\right) + 1 - 15 = -\frac{1}{9} + \frac{5}{3} + 1 - 15 = -\frac{1}{9} + \frac{15}{9} + 1 - 15 = \frac{14}{9} + 1 - 15 = \frac{23}{9} - 15 = \frac{23}{9} - \frac{135}{9} = -\frac{112}{9} \neq 0 \][/tex]
Since [tex]\( P\left(\frac{1}{3}\right) \neq 0 \)[/tex], [tex]\( 3x - 1 \)[/tex] is not a factor.
3. Factor: [tex]\( x + 5 \)[/tex]
- To test if [tex]\( x + 5 \)[/tex] is a factor, substitute [tex]\( x = -5 \)[/tex] into the polynomial:
[tex]\[ P(-5) = -3(-5)^3 + 15(-5)^2 + 3(-5) - 15 = -3(-125) + 15(25) - 15 - 15 = 375 + 375 - 15 - 15 = 720 \neq 0 \][/tex]
Since [tex]\( P(-5) \neq 0 \)[/tex], [tex]\( x + 5 \)[/tex] is not a factor.
4. Factor: [tex]\( -3 \)[/tex]
- Since [tex]\(-3\)[/tex] is a numeric constant (a factor that cannot be zero as a polynomial zero), it cannot be considered a suitable factor in this context. Factors of a polynomial in [tex]\( x \)[/tex] should involve [tex]\( x \)[/tex].
5. Factor: [tex]\( x - 5 \)[/tex]
- To test if [tex]\( x - 5 \)[/tex] is a factor, substitute [tex]\( x = 5 \)[/tex] into the polynomial:
[tex]\[ P(5) = -3(5)^3 + 15(5)^2 + 3(5) - 15 = -3(125) + 15(25) + 15 - 15 = -375 + 375 + 15 - 15 = 0 \][/tex]
Since [tex]\( P(5) = 0 \)[/tex], [tex]\( x - 5 \)[/tex] is a factor.
6. Factor: [tex]\( x - 1 \)[/tex]
- To test if [tex]\( x - 1 \)[/tex] is a factor, substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ P(1) = -3(1)^3 + 15(1)^2 + 3(1) - 15 = -3(1) + 15(1) + 3 - 15 = -3 + 15 + 3 - 15 = 0 \][/tex]
Since [tex]\( P(1) = 0 \)[/tex], [tex]\( x - 1 \)[/tex] is a factor.
7. Factor: [tex]\( 3x \)[/tex]
- For [tex]\( 3x \)[/tex] to be a factor, the polynomial should be zero when [tex]\( x = 0 \)[/tex]:
[tex]\[ P(0) = -3(0)^3 + 15(0)^2 + 3(0) - 15 = -15 \neq 0 \][/tex]
Since [tex]\( P(0) \neq 0 \)[/tex], [tex]\( 3x \)[/tex] is not a factor.
In conclusion, the correct factors of the polynomial [tex]\( -3 x^3 + 15 x^2 + 3 x - 15 \)[/tex] among the given options are:
- [tex]\( x - 5 \)[/tex]
- [tex]\( x - 1 \)[/tex]
First, write the polynomial in question:
[tex]\[ P(x) = -3 x^3 + 15 x^2 + 3 x - 15 \][/tex]
Now let's test each of the given possible factors to see which ones are actual factors of the polynomial.
1. Factor: [tex]\( \pi + 1 \)[/tex]
- This is a constant and not a suitable factor for a polynomial in [tex]\( x \)[/tex]. Factors of a polynomial should also be polynomial expressions in [tex]\( x \)[/tex].
2. Factor: [tex]\( 3x - 1 \)[/tex]
- To test if [tex]\( 3x - 1 \)[/tex] is a factor, substitute [tex]\( x = \frac{1}{3} \)[/tex] into the polynomial:
[tex]\[ P\left(\frac{1}{3}\right) = -3 \left(\frac{1}{3}\right)^3 + 15 \left(\frac{1}{3}\right)^2 + 3 \left(\frac{1}{3}\right) - 15 = -3 \left(\frac{1}{27}\right) + 15 \left(\frac{1}{9}\right) + 1 - 15 = -\frac{1}{9} + \frac{5}{3} + 1 - 15 = -\frac{1}{9} + \frac{15}{9} + 1 - 15 = \frac{14}{9} + 1 - 15 = \frac{23}{9} - 15 = \frac{23}{9} - \frac{135}{9} = -\frac{112}{9} \neq 0 \][/tex]
Since [tex]\( P\left(\frac{1}{3}\right) \neq 0 \)[/tex], [tex]\( 3x - 1 \)[/tex] is not a factor.
3. Factor: [tex]\( x + 5 \)[/tex]
- To test if [tex]\( x + 5 \)[/tex] is a factor, substitute [tex]\( x = -5 \)[/tex] into the polynomial:
[tex]\[ P(-5) = -3(-5)^3 + 15(-5)^2 + 3(-5) - 15 = -3(-125) + 15(25) - 15 - 15 = 375 + 375 - 15 - 15 = 720 \neq 0 \][/tex]
Since [tex]\( P(-5) \neq 0 \)[/tex], [tex]\( x + 5 \)[/tex] is not a factor.
4. Factor: [tex]\( -3 \)[/tex]
- Since [tex]\(-3\)[/tex] is a numeric constant (a factor that cannot be zero as a polynomial zero), it cannot be considered a suitable factor in this context. Factors of a polynomial in [tex]\( x \)[/tex] should involve [tex]\( x \)[/tex].
5. Factor: [tex]\( x - 5 \)[/tex]
- To test if [tex]\( x - 5 \)[/tex] is a factor, substitute [tex]\( x = 5 \)[/tex] into the polynomial:
[tex]\[ P(5) = -3(5)^3 + 15(5)^2 + 3(5) - 15 = -3(125) + 15(25) + 15 - 15 = -375 + 375 + 15 - 15 = 0 \][/tex]
Since [tex]\( P(5) = 0 \)[/tex], [tex]\( x - 5 \)[/tex] is a factor.
6. Factor: [tex]\( x - 1 \)[/tex]
- To test if [tex]\( x - 1 \)[/tex] is a factor, substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ P(1) = -3(1)^3 + 15(1)^2 + 3(1) - 15 = -3(1) + 15(1) + 3 - 15 = -3 + 15 + 3 - 15 = 0 \][/tex]
Since [tex]\( P(1) = 0 \)[/tex], [tex]\( x - 1 \)[/tex] is a factor.
7. Factor: [tex]\( 3x \)[/tex]
- For [tex]\( 3x \)[/tex] to be a factor, the polynomial should be zero when [tex]\( x = 0 \)[/tex]:
[tex]\[ P(0) = -3(0)^3 + 15(0)^2 + 3(0) - 15 = -15 \neq 0 \][/tex]
Since [tex]\( P(0) \neq 0 \)[/tex], [tex]\( 3x \)[/tex] is not a factor.
In conclusion, the correct factors of the polynomial [tex]\( -3 x^3 + 15 x^2 + 3 x - 15 \)[/tex] among the given options are:
- [tex]\( x - 5 \)[/tex]
- [tex]\( x - 1 \)[/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
