To find the real zeros of the polynomial f(x) = x³ + 4x² + 9x + 36, we need to determine the values of x for which f(x) = 0.
The first step is to look for rational roots using the Rational Root Theorem, which suggests that any possible rational root, in this case, would be a factor of the constant term (36) divided by a factor of the leading coefficient (1). The factors of 36 are: ±1, ±2, ±3, ±4, ±9, ±12, ±18, ±36.
We can test these root candidates by substituting them into the polynomial. Let’s evaluate:
- f(1) = 1³ + 4(1)² + 9(1) + 36 = 1 + 4 + 9 + 36 = 50
- f(-1) = (-1)³ + 4(-1)² + 9(-1) + 36 = -1 + 4 – 9 + 36 = 30
- f(2) = 2³ + 4(2)² + 9(2) + 36 = 8 + 16 + 18 + 36 = 78
- f(-2) = (-2)³ + 4(-2)² + 9(-2) + 36 = -8 + 16 – 18 + 36 = 26
- f(3) = 3³ + 4(3)² + 9(3) + 36 = 27 + 36 + 27 + 36 = 126
- f(-3) = (-3)³ + 4(-3)² + 9(-3) + 36 = -27 + 36 – 27 + 36 = 18
- f(4) = 4³ + 4(4)² + 9(4) + 36 = 64 + 64 + 36 + 36 = 200
- f(-4) = (-4)³ + 4(-4)² + 9(-4) + 36 = -64 + 64 – 36 + 36 = 0
We’ve found that f(-4) = 0, meaning x = -4 is a real root. Next, we can use polynomial long division or synthetic division to divide the original polynomial by (x + 4).
After dividing x³ + 4x² + 9x + 36 by x + 4, we find the quotient is x² + 0x + 9, which simplifies to x² + 9.
The next step is to find the zeros of the quotient. Setting x² + 9 = 0 gives:
x² = -9
x = ±√(-9) = ±3i
Since 3i and -3i are not real numbers, we only have the real zero from our initial factorization.
In summary, the only real zero of the polynomial x³ + 4x² + 9x + 36 is:
- x = -4
This indicates that the polynomial crosses the x-axis at the point (-4, 0), while the other zeros are complex.