To find the real zeros of the polynomial x³ + 4x² + 9x + 36, we first set the polynomial equal to zero:
x³ + 4x² + 9x + 36 = 0
Next, we can look for possible rational roots using the Rational Root Theorem which suggests that any rational solution is a factor of the constant term (36) divided by a factor of the leading coefficient (1). The factors of 36 include ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
Testing some of these values: we can start with x = -2:
(-2)³ + 4(-2)² + 9(-2) + 36 = -8 + 16 – 18 + 36 = 26 (not a root)
Next, we’ll try x = -3:
(-3)³ + 4(-3)² + 9(-3) + 36 = -27 + 36 – 27 + 36 = 18 (not a root)
Then, let’s check x = -4:
(-4)³ + 4(-4)² + 9(-4) + 36 = -64 + 64 – 36 + 36 = 0 (which means x = -4 is a root)
Now that we’ve found a root, we can perform polynomial long division on x³ + 4x² + 9x + 36 by (x + 4) to factor the polynomial further:
After performing the division, we find:
x³ + 4x² + 9x + 36 = (x + 4)(x² + 9)
The remaining factor x² + 9 does not have any real roots, as it can be set to zero:
x² + 9 = 0 ➜ x² = -9
This gives us complex roots x = ±3i (which are not real).
Therefore, the only real zero of the polynomial x³ + 4x² + 9x + 36 is:
x = -4.