Answer :
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]
