Answer:
To find the zeros of the function f(x) = 4x^3+44x^2+152x+160, we can use one of several methods. One possible method is as follows:
Step 1: Factor out the common factor 4 from all terms:
f(x) = 4(x^3+11x^2+38x+40)
Step 2: Use the rational root theorem to find possible rational zeros. The rational root theorem states that any rational zero of a polynomial with integer coefficients can be written as a fraction p/q, where p is a factor of the constant term (in this case, 40) and q is a factor of the leading coefficient (in this case, 4).
The factors of 40 are ±1, ±2, ±4, ±5, ±8, ±10, ±20, and ±40.
The factors of 4 are ±1 and ±2.
Therefore, the possible rational zeros are:
±1/4, ±1/2, ±1, ±2, ±4, ±5/2, ±10, and ±20.
Step 3: Test each possible rational zero using synthetic division, until a zero is found. For example, using synthetic division with a possible zero of -4, we get:
-4 | 4 44 152 160
| -16 -112 -160
-----------------
4 28 40 0
Since the remainder is zero, -4 is a zero of f(x), and we have successfully factored f(x) as:
f(x) = 4(x+4)(x^2+7x+10)
Step 4: Solve for the remaining zeros of the quadratic factor x^2+7x+10 using the quadratic formula:
x = [-b ± sqrt(b^2-4ac)] / 2a
= [-7 ± sqrt(7^2-4(1)(10))] / 2
= (-7 ± 3) / 2
Therefore, the zeros of f(x) are:
x = -4, -5, -2