Find all the zeros of the function.
f(x) = x^3 + 11x^2 + 39x + 29
I found one of the zeros = -1 but I am confused on how to find the other two.
As suggested by Mr. Pursley, from the known root, do a long division to arrive at the quotient (x^2+10*x+29) from which you can solve by the quadratic formula to get the two remaining (complex) roots, namely -5-2i and -5+2i.
To find all the zeros of the given function f(x), you need to factor it completely.
Step 1: Start by checking the known zero, which is x = -1. Substitute this value into the function and check if it equals zero:
f(-1) = (-1)^3 + 11(-1)^2 + 39(-1) + 29
= -1 + 11 - 39 + 29
= 0
Since f(-1) = 0, we have confirmed that x = -1 is a zero of the function.
Step 2: Now, you need to perform synthetic division to divide f(x) by (x + 1). This will give you a quadratic equation for which you can find the remaining two zeros. The coefficient of x^3 is 1, so we will divide each term by 1.
Performing synthetic division:
-1 | 1 11 39 29
|___________
| -1 10 -49
___________
1 10 -49 -20
The result after dividing f(x) = x^3 + 11x^2 + 39x + 29 by (x + 1) is x^2 + 10x - 49 with a remainder of -20. So, the quadratic equation is x^2 + 10x - 49.
Step 3: Now, solve the quadratic equation x^2 + 10x - 49 = 0 to find the remaining two zeros. You can use factoring, completing the square, or the quadratic formula.
Factoring:
(x + 7)(x - 7) = 0
Setting each factor equal to zero:
x + 7 = 0 or x - 7 = 0
Solving each equation:
x = -7 or x = 7
Now, you have found all three zeros of the function f(x).
The zeros are: -1, -7, and 7.