Answer :
To find the real solutions of the polynomial equation [tex]\( 3x^3 + 4x^2 - 7x + 2 = 0 \)[/tex], we can use various methods such as factoring, the Rational Root Theorem, or numerical techniques. We'll start by using the Rational Root Theorem to look for any possible rational solutions.
The Rational Root Theorem states that any rational root of a polynomial equation (with integer coefficients) must be a factor of the constant term divided by a factor of the leading coefficient. For our equation:
- The constant term is [tex]\(2\)[/tex].
- The leading coefficient is [tex]\(3\)[/tex].
The possible rational roots are therefore [tex]\( \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3} \)[/tex].
We will test these possible rational roots in the polynomial to see if any of them satisfy the equation:
1. Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ 3(1)^3 + 4(1)^2 - 7(1) + 2 = 3 + 4 - 7 + 2 = 2 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
2. Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ 3(-1)^3 + 4(-1)^2 - 7(-1) + 2 = -3 + 4 + 7 + 2 = 10 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
3. Testing [tex]\( x = 2 \)[/tex]:
[tex]\[ 3(2)^3 + 4(2)^2 - 7(2) + 2 = 24 + 16 - 14 + 2 = 28 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
4. Testing [tex]\( x = -2 \)[/tex]:
[tex]\[ 3(-2)^3 + 4(-2)^2 - 7(-2) + 2 = -24 + 16 + 14 + 2 = 8 \neq 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is not a root.
5. Testing [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ 3\left(\frac{1}{3}\right)^3 + 4\left(\frac{1}{3}\right)^2 - 7\left(\frac{1}{3}\right) + 2 = \frac{3}{27} + \frac{4}{9} - \frac{7}{3} + 2 = \frac{1}{9} + \frac{4}{9} - \frac{21}{9} + 2 = -\frac{16}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{3} \)[/tex] is not a root.
6. Testing [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ 3\left(-\frac{1}{3}\right)^3 + 4\left(-\frac{1}{3}\right)^2 - 7\left(-\frac{1}{3}\right) + 2 = -\frac{3}{27} + \frac{4}{9} + \frac{7}{3} + 2 = -\frac{1}{9} + \frac{4}{9} + \frac{21}{9} + 2 = \frac{24}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = -\frac{1}{3} \)[/tex] is not a root.
7. Testing [tex]\( x = \frac{2}{3} \)[/tex]:
[tex]\[ 3\left(\frac{2}{3}\right)^3 + 4\left(\frac{2}{3}\right)^2 - 7\left(\frac{2}{3}\right) + 2 = \frac{3 \cdot 8}{27} + \frac{4 \cdot 4}{9} - \frac{14}{3} + 2 = \frac{8}{9} + \frac{16}{9} - \frac{42}{9} + 2 = -\frac{18}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = \frac{2}{3} \)[/tex] is not a root.
8. Testing [tex]\( x = -\frac{2}{3} \)[/tex]:
[tex]\[ 3\left(-\frac{2}{3}\right)^3 + 4\left(-\frac{2}{3}\right)^2 - 7\left(-\frac{2}{3}\right) + 2 = \frac{3(-8)}{27} + \frac{4(4)}{9} + \frac{14}{3} + 2 = -\frac{8}{9} + \frac{16}{9} + \frac{42}{9} + 2 = \frac{50}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = -\frac{2}{3} \)[/tex] is not a root.
Since none of these are roots, there are likely no rational solutions. At this point, to find the real roots, we can turn to numerical methods or a graphing calculator, but here we will assume the problem is done analytically.
Since all rational root tests have failed, we conclude:
The solution set is Q.
So, our answer is:
B. The solution set is [tex]\( Q \)[/tex].
The Rational Root Theorem states that any rational root of a polynomial equation (with integer coefficients) must be a factor of the constant term divided by a factor of the leading coefficient. For our equation:
- The constant term is [tex]\(2\)[/tex].
- The leading coefficient is [tex]\(3\)[/tex].
The possible rational roots are therefore [tex]\( \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3} \)[/tex].
We will test these possible rational roots in the polynomial to see if any of them satisfy the equation:
1. Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ 3(1)^3 + 4(1)^2 - 7(1) + 2 = 3 + 4 - 7 + 2 = 2 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
2. Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ 3(-1)^3 + 4(-1)^2 - 7(-1) + 2 = -3 + 4 + 7 + 2 = 10 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
3. Testing [tex]\( x = 2 \)[/tex]:
[tex]\[ 3(2)^3 + 4(2)^2 - 7(2) + 2 = 24 + 16 - 14 + 2 = 28 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
4. Testing [tex]\( x = -2 \)[/tex]:
[tex]\[ 3(-2)^3 + 4(-2)^2 - 7(-2) + 2 = -24 + 16 + 14 + 2 = 8 \neq 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is not a root.
5. Testing [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ 3\left(\frac{1}{3}\right)^3 + 4\left(\frac{1}{3}\right)^2 - 7\left(\frac{1}{3}\right) + 2 = \frac{3}{27} + \frac{4}{9} - \frac{7}{3} + 2 = \frac{1}{9} + \frac{4}{9} - \frac{21}{9} + 2 = -\frac{16}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{3} \)[/tex] is not a root.
6. Testing [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ 3\left(-\frac{1}{3}\right)^3 + 4\left(-\frac{1}{3}\right)^2 - 7\left(-\frac{1}{3}\right) + 2 = -\frac{3}{27} + \frac{4}{9} + \frac{7}{3} + 2 = -\frac{1}{9} + \frac{4}{9} + \frac{21}{9} + 2 = \frac{24}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = -\frac{1}{3} \)[/tex] is not a root.
7. Testing [tex]\( x = \frac{2}{3} \)[/tex]:
[tex]\[ 3\left(\frac{2}{3}\right)^3 + 4\left(\frac{2}{3}\right)^2 - 7\left(\frac{2}{3}\right) + 2 = \frac{3 \cdot 8}{27} + \frac{4 \cdot 4}{9} - \frac{14}{3} + 2 = \frac{8}{9} + \frac{16}{9} - \frac{42}{9} + 2 = -\frac{18}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = \frac{2}{3} \)[/tex] is not a root.
8. Testing [tex]\( x = -\frac{2}{3} \)[/tex]:
[tex]\[ 3\left(-\frac{2}{3}\right)^3 + 4\left(-\frac{2}{3}\right)^2 - 7\left(-\frac{2}{3}\right) + 2 = \frac{3(-8)}{27} + \frac{4(4)}{9} + \frac{14}{3} + 2 = -\frac{8}{9} + \frac{16}{9} + \frac{42}{9} + 2 = \frac{50}{9} + 2 \neq 0 \][/tex]
So, [tex]\( x = -\frac{2}{3} \)[/tex] is not a root.
Since none of these are roots, there are likely no rational solutions. At this point, to find the real roots, we can turn to numerical methods or a graphing calculator, but here we will assume the problem is done analytically.
Since all rational root tests have failed, we conclude:
The solution set is Q.
So, our answer is:
B. The solution set is [tex]\( Q \)[/tex].
