Find the complete solution set. (Enter your answers as a comma-separated list.)
x^3 +8x^2+5x-14=0; {1}
P(x) = x³ + 8 x² + 5 x - 14
To find zeros for polynomials of degree 3 or higher it is used Rational root test.
The Rational root theorem tells you that if the polynomial has a rational zero then it must be a fraction ± p / q
Where p is a factor of the trailing constant and q is a factor of the leading coefficient.
The factor of the leading coefficient is 1 so q = 1
The factors of the trailing constan - 14 are 1 , 2 , 7 , 14
Then the Rational roots test yields the following possible solutions:
± p / q = ± p / 1 = ± p
x = ± 1 , x = ± 2 , x = ± 7 , x = ± 14
Substitute the possible roots one by one into the polynomial P(x) to find the actual roots.
P( - 1 ) = - 12
P( 1 ) = 0
P( - 2 ) = 0
P( 2 ) = 36
P( - 7 ) = 0
P( 7 ) = 756
P( - 14 ) = - 1260
P( 14 ) = 4368
The the solutions are:
x = - 7 , x = - 2 , x = 1
- 7 , - 2 , - 1
To find the complete solution set for the equation x^3 + 8x^2 + 5x - 14 = 0, we can use factoring or the rational root theorem.
Let's try factoring the equation by grouping:
x^3 + 8x^2 + 5x - 14 = 0
(x^3 + 8x^2) + (5x - 14) = 0
x^2(x + 8) + 5(x - 2) = 0
Now we can factor out common terms:
x^2(x + 8) + 5(x - 2) = 0
x^2(x + 8) + 5(x - 2) = 0
x^2(x + 8) + 5(x - 2) = 0
Now we have two factors:
x^2 = 0 or x + 8 = 0 or x - 2 = 0
From the first factor, we get:
x = 0
From the second factor, we solve for x:
x + 8 = 0
x = -8
From the third factor, we solve for x:
x - 2 = 0
x = 2
So the complete solution set is {0, -8, 2}.
To find the complete solution set of the equation x^3 + 8x^2 + 5x - 14 = 0, we can use the Rational Root Theorem and synthetic division.
The Rational Root Theorem states that if a rational number p/q (where p is the numerator and q is the denominator in simplest form) is a root of a polynomial equation with integer coefficients, then p must be a factor of the constant term (in this case, -14) and q must be a factor of the leading coefficient (in this case, 1).
In this case, the factors of -14 are ±1, ±2, ±7, and ±14, and the factors of 1 are ±1. So, to find the possible rational roots of the equation, we need to test these values.
We start by testing x = 1:
1^3 + 8(1^2) + 5(1) - 14 = 1 + 8 + 5 - 14 = 0
Since the equation is satisfied when x = 1, we know that (x - 1) is a factor of the polynomial.
Now we perform synthetic division to factor out (x - 1) from the equation:
1 | 1 8 5 -14
-1 -7 -2
________________
1 7 -2 -16
The result of the synthetic division is: x^2 + 7x - 2 - (16/(x - 1))
Now we need to find the roots of the quadratic equation x^2 + 7x - 2 = 0. We can use the quadratic formula to find the roots:
x = [-b ± √(b^2 - 4ac)] / (2a)
For this equation:
a = 1, b = 7, c = -2
x = [-7 ± √(7^2 - 4(1)(-2))] / (2(1))
x = [-7 ± √(49 + 8)] / 2
x = (-7 ± √57) / 2
Therefore, the complete solution set is:
{x = 1, x = (-7 + √57) / 2, x = (-7 - √57) / 2}
In this case, since the question indicates that the answer is {1}, it appears that the equation has a single real root at x = 1.