Answer :
To find all roots (real and non-real) of the polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex], follow these steps:
### Step 1: Identify the Polynomial
The polynomial given is
[tex]\[ p(x) = 2x^3 - 3x^2 + 32x + 17. \][/tex]
### Step 2: Look for Rational Roots
First, use the Rational Root Theorem to test for possible rational roots. The Rational Root Theorem states that any possible rational root, [tex]\( p/q \)[/tex], is such that [tex]\( p \)[/tex] is a factor of the constant term (17) and [tex]\( q \)[/tex] is a factor of the leading coefficient (2).
Factors of 17: [tex]\( \pm 1, \pm 17 \)[/tex]
Factors of 2: [tex]\( \pm 1, \pm 2 \)[/tex]
Possible rational roots are:
[tex]\[ \pm 1, \pm 17, \pm \frac{1}{2}, \pm \frac{17}{2} \][/tex]
### Step 3: Test Rational Roots
Substitute these potential rational roots into the polynomial to see if any of them result in zero.
Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 2(1)^3 - 3(1)^2 + 32(1) + 17 = 2 - 3 + 32 + 17 = 48 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ p(-1) = 2(-1)^3 - 3(-1)^2 + 32(-1) + 17 = -2 - 3 - 32 + 17 = -20 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
Continuing this process for other potential rational roots, you will find that none of the rational roots make the polynomial equal zero.
### Step 4: Use Algebraic Methods to Find Complex Roots
Since the rational root approach didn't yield any real roots, we move to solving the polynomial for its complex roots.
The cubic polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] can be tricky to solve by algebraic methods or factoring, so we use techniques such as the cubic formula or numerical methods for finding roots.
### Step 5: Solve the Polynomial
The roots of the polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] are found to be:
[tex]\[ x = -\frac{1}{2}, \quad x = 1 - 4i, \quad \text{and} \quad x = 1 + 4i. \][/tex]
### Step 6: Verification
These solutions can be verified by substituting each root back into the polynomial:
For [tex]\( x = -\frac{1}{2} \)[/tex]:
[tex]\[ 2 \left( -\frac{1}{2} \right)^3 - 3 \left( -\frac{1}{2} \right)^2 + 32 \left( -\frac{1}{2} \right) + 17 = 0 \][/tex]
For [tex]\( x = 1 - 4i \)[/tex]:
[tex]\[ 2(1 - 4i)^3 - 3(1 - 4i)^2 + 32(1 - 4i) + 17 = 0 \][/tex]
For [tex]\( x = 1 + 4i \)[/tex]:
[tex]\[ 2(1 + 4i)^3 - 3(1 + 4i)^2 + 32(1 + 4i) + 17 = 0 \][/tex]
Therefore, the complete set of solutions, including both real and non-real, of the polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] are:
[tex]\[ x = -\frac{1}{2}, \quad x = 1 - 4i, \quad \text{and} \quad x = 1 + 4i. \][/tex]
### Step 1: Identify the Polynomial
The polynomial given is
[tex]\[ p(x) = 2x^3 - 3x^2 + 32x + 17. \][/tex]
### Step 2: Look for Rational Roots
First, use the Rational Root Theorem to test for possible rational roots. The Rational Root Theorem states that any possible rational root, [tex]\( p/q \)[/tex], is such that [tex]\( p \)[/tex] is a factor of the constant term (17) and [tex]\( q \)[/tex] is a factor of the leading coefficient (2).
Factors of 17: [tex]\( \pm 1, \pm 17 \)[/tex]
Factors of 2: [tex]\( \pm 1, \pm 2 \)[/tex]
Possible rational roots are:
[tex]\[ \pm 1, \pm 17, \pm \frac{1}{2}, \pm \frac{17}{2} \][/tex]
### Step 3: Test Rational Roots
Substitute these potential rational roots into the polynomial to see if any of them result in zero.
Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 2(1)^3 - 3(1)^2 + 32(1) + 17 = 2 - 3 + 32 + 17 = 48 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ p(-1) = 2(-1)^3 - 3(-1)^2 + 32(-1) + 17 = -2 - 3 - 32 + 17 = -20 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
Continuing this process for other potential rational roots, you will find that none of the rational roots make the polynomial equal zero.
### Step 4: Use Algebraic Methods to Find Complex Roots
Since the rational root approach didn't yield any real roots, we move to solving the polynomial for its complex roots.
The cubic polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] can be tricky to solve by algebraic methods or factoring, so we use techniques such as the cubic formula or numerical methods for finding roots.
### Step 5: Solve the Polynomial
The roots of the polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] are found to be:
[tex]\[ x = -\frac{1}{2}, \quad x = 1 - 4i, \quad \text{and} \quad x = 1 + 4i. \][/tex]
### Step 6: Verification
These solutions can be verified by substituting each root back into the polynomial:
For [tex]\( x = -\frac{1}{2} \)[/tex]:
[tex]\[ 2 \left( -\frac{1}{2} \right)^3 - 3 \left( -\frac{1}{2} \right)^2 + 32 \left( -\frac{1}{2} \right) + 17 = 0 \][/tex]
For [tex]\( x = 1 - 4i \)[/tex]:
[tex]\[ 2(1 - 4i)^3 - 3(1 - 4i)^2 + 32(1 - 4i) + 17 = 0 \][/tex]
For [tex]\( x = 1 + 4i \)[/tex]:
[tex]\[ 2(1 + 4i)^3 - 3(1 + 4i)^2 + 32(1 + 4i) + 17 = 0 \][/tex]
Therefore, the complete set of solutions, including both real and non-real, of the polynomial [tex]\( 2x^3 - 3x^2 + 32x + 17 \)[/tex] are:
[tex]\[ x = -\frac{1}{2}, \quad x = 1 - 4i, \quad \text{and} \quad x = 1 + 4i. \][/tex]
