Find all the real zeros of the polynomial and use the quadratic formula. X^4-7x^3+14x^2-3x-9
To find the real zeros of the polynomial, we can use the Rational Root Theorem. The possible rational roots of the polynomial can be found by taking the factors of the constant term (-9) divided by the factors of the leading coefficient (1).
The factors of the constant term (-9) are ±1, ±3, ±9.
The factors of the leading coefficient (1) are ±1.
So, the possible rational roots are:
±1, ±3, ±9.
To check which of these possible rational roots are actual roots of the polynomial, we can substitute each of them into the polynomial and see if the result is equal to zero.
Let's start with 1:
(1)^4 - 7(1)^3 + 14(1)^2 - 3(1) - 9 = 1 - 7 + 14 - 3 - 9 = -4.
Since the result is not equal to zero, 1 is not a zero of the polynomial.
Now let's try -1:
(-1)^4 - 7(-1)^3 + 14(-1)^2 - 3(-1) - 9 = 1 + 7 + 14 + 3 - 9 = 16.
Since the result is not equal to zero, -1 is not a zero of the polynomial.
Let's move on to 3:
(3)^4 - 7(3)^3 + 14(3)^2 - 3(3) - 9 = 81 - 189 + 126 - 9 - 9 = 0.
Since the result is equal to zero, 3 is a zero of the polynomial.
Therefore, the polynomial has a real zero at x = 3.
To find the other zeros of the polynomial, we can divide the polynomial by (x - 3) using synthetic division:
3 | 1 -7 14 -3 -9
| 3 -12 6 9
-----------------
1 -4 2 3 0
The quotient is 1x^3 - 4x^2 + 2x + 3, which is a cubic polynomial. To find the remaining zeros, we can apply the same process again.
Now let's find the possible rational roots for the cubic polynomial:
The factors of the constant term (0) are ±1, ±3.
The factors of the leading coefficient (1) are ±1.
So, the possible rational roots are:
±1, ±3.
We can try each of these possible rational roots to check if they are actual zeros of the cubic polynomial.
Let's try 1:
(1)^3 - 4(1)^2 + 2(1) + 3 = 1 - 4 + 2 + 3 = 2.
Since the result is not equal to zero, 1 is not a zero of the cubic polynomial.
Now let's try -1:
(-1)^3 - 4(-1)^2 + 2(-1) + 3 = -1 - 4 - 2 + 3 = -4.
Since the result is not equal to zero, -1 is not a zero of the cubic polynomial.
Now let's try 3:
(3)^3 - 4(3)^2 + 2(3) + 3 = 27 - 36 + 6 + 3 = 0.
Since the result is equal to zero, 3 is a zero of the cubic polynomial.
Therefore, the polynomial has a real zero at x = 3.
To find the remaining zeros, we can divide the cubic polynomial by (x - 3) using synthetic division:
3 | 1 -4 2 3
| 3 -3 -3
---------------
1 -1 -1 0
The quotient is 1x^2 - x - 1, which is a quadratic polynomial.
Now we can use the quadratic formula to find the remaining zeros.
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic polynomial 1x^2 - x - 1,
a = 1, b = -1, c = -1.
x = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
= (1 ± √(1 + 4)) / 2
= (1 ± √5) / 2
Therefore, the remaining zeros of the polynomial are:
x = (1 + √5) / 2
x = (1 - √5) / 2
In summary, the real zeros of the polynomial are: x = 3, (1 + √5) / 2, and (1 - √5) / 2.