Find the roots of the polynomial equation. x^3+6x^2+11x+12=0
To find the roots of the given polynomial equation x^3 + 6x^2 + 11x + 12 = 0, we can use the Rational Root Theorem to check for possible rational roots.
The Rational Root Theorem states that if a polynomial equation has a rational root p/q, where p is the factor of the constant term (12) and q is the factor of the leading coefficient (1), then p must be a divisor of 12 and q must be a divisor of 1.
The factors of 12 are {±1, ±2, ±3, ±4, ±6, ±12}, and the factors of 1 are {±1}. Therefore, the possible rational roots are:
±1, ±2, ±3, ±4, ±6, ±12.
Let's test these values and see if any of them satisfy the equation:
For x = -1:
(-1)^3 + 6(-1)^2 + 11(-1) + 12 = -1 + 6 - 11 + 12 = 6
For x = 1:
(1)^3 + 6(1)^2 + 11(1) + 12 = 1 + 6 + 11 + 12 = 30
For x = -2:
(-2)^3 + 6(-2)^2 + 11(-2) + 12 = -8 + 24 - 22 + 12 = 6
For x = 2:
(2)^3 + 6(2)^2 + 11(2) + 12 = 8 + 24 + 22 + 12 = 66
For x = -3:
(-3)^3 + 6(-3)^2 + 11(-3) + 12 = -27 + 54 - 33 + 12 = 6
For x = 3:
(3)^3 + 6(3)^2 + 11(3) + 12 = 27 + 54 + 33 + 12 = 126
For x = -4:
(-4)^3 + 6(-4)^2 + 11(-4) + 12 = -64 + 96 - 44 + 12 = 0
Therefore, x = -4 is a root of the polynomial equation x^3 + 6x^2 + 11x + 12 = 0.
Using synthetic division with x = -4 as the root, we divide the polynomial by (x + 4):
-4 | 1 6 11 12
-4 -8 -12
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1 2 -1 0
The remainder is 0, and the quotient is 1x^2 + 2x - 1.
Now let's solve the quadratic equation 1x^2 + 2x - 1 = 0 to find the remaining roots:
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, where a = 1, b = 2, and c = -1:
x = (-2 ± √(2^2 - 4(1)(-1))) / (2(1))
x = (-2 ± √(4 + 4)) / 2
x = (-2 ± √8) / 2
x = (-2 ± 2√2) / 2
x = -1 ± √2
Therefore, the roots of the polynomial equation x^3 + 6x^2 + 11x + 12 = 0 are -4, -1 + √2, and -1 - √2.