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
Find the complete solution set. (Enter your answers as a comma-separated list.)
x^3 +8x^2+5x-14=0; {1}
1 answer
1 answer