To factor the polynomial f(x) = x³ + 2x² + 5x + 6, we start by trying to find its roots using the Rational Root Theorem, which suggests that any rational root, in the form p/q, is a factor of the constant term 6 divided by a factor of the leading coefficient (which is 1 here).
The potential rational roots of the polynomial are ±1, ±2, ±3, ±6. We can test these values by substituting them into the polynomial to see if any yield zero.
– Testing x = -1:
f(-1) = (-1)³ + 2(-1)² + 5(-1) + 6 = -1 + 2 – 5 + 6 = 2 (not a root)
– Testing x = -2:
f(-2) = (-2)³ + 2(-2)² + 5(-2) + 6 = -8 + 8 – 10 + 6 = -4 (not a root)
– Testing x = -3:
f(-3) = (-3)³ + 2(-3)² + 5(-3) + 6 = -27 + 18 – 15 + 6 = -18 (not a root)
– Testing x = -2 again shows that we might have specific factors we can look for:
Now, let’s group and factor the expression differently. We can factor by grouping:
1. Take the first two terms: x³ + 2x² = x²(x + 2)
2. Take the last two terms: 5x + 6 = 5(x + 2)
Now combine these:
f(x) = x²(x + 2) + 5(x + 2)
Notice that both groups have a common factor of (x + 2):
f(x) = (x + 2)(x² + 5)
Your completely factored form of the polynomial f(x) = x³ + 2x² + 5x + 6 is:
(x + 2)(x² + 5)
Here, (x + 2) is a linear factor, and (x² + 5) cannot be factored further over the real numbers since it has no real roots (its discriminant is negative).