To factor the polynomial x² + 6x + 27, we will apply the methods of factoring polynomials. First, let’s check if it can be factored as a simple quadratic equation of the form ax² + bx + c.
In our case:
- a = 1
- b = 6
- c = 27
To find the factors of this polynomial, we can look for two numbers that multiply to ac = 1 * 27 = 27 and add up to b = 6. However, upon examining the pairs of factors of 27, we find:
- (1, 27)
- (3, 9)
None of these pairs add up to 6. This indicates that x² + 6x + 27 cannot be factored using integer values. To verify this, we can also compute the discriminant of the quadratic equation.
The discriminant D of the quadratic equation is given by:
D = b² – 4ac
Substituting the values:
D = 6² – 4 * 1 * 27 = 36 – 108 = -72
Since the discriminant is negative (D < 0), this confirms that there are no real roots, which means that x² + 6x + 27 cannot be factored into real linear factors.
However, if we are looking for a different kind of ‘factored’ form, we can express it using complex numbers. The roots of this polynomial can be found using the quadratic formula:
x = (-b ± √D) / 2a
Substituting our values:
x = (-6 ± √(-72)) / 2
Calculating the square root of -72 gives us:
√(-72) = 6i√2
Now we can write the roots:
x = (-6 ± 6i√2) / 2 = -3 ± 3i√2
This means the factored form using complex roots can be expressed as:
(x + 3 – 3i√2)(x + 3 + 3i√2)
In conclusion, the polynomial x² + 6x + 27 does not have a factored form using real numbers, but its factored form using complex numbers is:
(x + 3 – 3i√2)(x + 3 + 3i√2)
This illustrates the importance of understanding both real and complex number factorizations in algebra.