1)Use synthetic subsitution to find f(-3)for f(x)=x^4-4x^3+2x^2-4x+6

answer= -15

2)One factor of x^3+2x^2-11x-12 is x+4. find the remaining factors
answer= x-1;x-3

3)Which describes the number and type of roots of the equation x^3+121x=0
answer= 1 real root,two imaginary roots

4)State the possible number of imaginary zeros of g(x)=x^4+3x^3+7x^2-6x-13
answer= 2,4,or 0

1. check your arithmetic, I got 225

2. I had (x-3)(x+1), (the answer after the division was x^2 - 2x -3

3. yes

4. there are two reals and two imaginary

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Apologies for the mistakes in the previous answers. Here are the corrected responses:

1) Using synthetic substitution to find f(-3) for f(x) = x^4 - 4x^3 + 2x^2 - 4x + 6:
By substituting -3 into the equation, we get (-3)^4 - 4(-3)^3 + 2(-3)^2 - 4(-3) + 6 = 81 + 108 + 18 + 12 + 6 = 225.

2) One factor of x^3 + 2x^2 - 11x - 12 is x + 4. To find the remaining factors, we can divide x^3 + 2x^2 - 11x - 12 by x + 4 using synthetic division:
Using synthetic division, we have:
-4 | 1 2 -11 -12
-4 8 12
---------------------
1 -2 -3 0

This means that x^3 + 2x^2 - 11x - 12 = (x + 4)(x^2 - 2x - 3).
Factoring x^2 - 2x - 3, we have:
x^3 + 2x^2 - 11x - 12 = (x + 4)(x - 1)(x - 3).

3) The equation x^3 + 121x = 0 has one real root and two imaginary roots. The real root is x = 0, and the imaginary roots are complex conjugates.

4) For g(x) = x^4 + 3x^3 + 7x^2 - 6x - 13, the possible number of imaginary zeros is either 2, 4, or 0.

1) To use synthetic substitution to find f(-3) for f(x) = x^4-4x^3+2x^2-4x+6, you will follow these steps:

Step 1: Replace every instance of x in the polynomial with the given value, which is -3 in this case.
f(-3) = (-3)^4 - 4(-3)^3 + 2(-3)^2 - 4(-3) + 6

Step 2: Simplify the expression using arithmetic.
f(-3) = 81 - 4(-27) + 2(9) + 12 + 6
f(-3) = 81 + 108 + 18 + 12 + 6
f(-3) = 225

Therefore, f(-3) = 225.

2) To find the remaining factors of x^3+2x^2-11x-12 given that x+4 is a factor, you can use synthetic division. Here's how:

Step 1: Set up the synthetic division table by placing the coefficients of the polynomial in descending order.
-4 | 1 2 -11 -12

Step 2: Bring down the first coefficient, which is 1, and multiply it by the divisor (-4) to get -4. Write this below the line.
-4 | 1 2 -11 -12
-4

Step 3: Add the value below the line (-4) to the next coefficient (2) to get -2. Multiply this sum by the divisor (-4) to get 8. Write this below the line.
-4 | 1 2 -11 -12
-4 8

Step 4: Repeat this process for the remaining coefficients.
-4 | 1 2 -11 -12
-4 8 12
______
1 -2 1 0

The resulting quotient is 1x^2 - 2x + 1, which can be factored as (x - 1)(x - 1) or (x - 1)^2.

Therefore, the remaining factors are x - 1 and x - 1 (or (x - 1)^2).

3) To determine the number and type of roots of the equation x^3 + 121x = 0, you can analyze the coefficients.

The equation has a degree of 3, so it will have 3 roots.

Since there is no constant term in the equation (the coefficient of x^0 is 0), it means that one of the roots is x = 0.

The other two roots can be found by setting the equation equal to 0 and solving for x:

x^3 + 121x = 0
x(x^2 + 121) = 0

Therefore, we have two cases:
1) x = 0: This is the real root.

2) x^2 + 121 = 0: Solving this equation, we get x = ±√(-121). Since the square root of a negative number is imaginary, we have two imaginary roots.

Hence, the number and type of roots of the equation x^3 + 121x = 0 are 1 real root and two imaginary roots.

4) To determine the possible number of imaginary zeros of g(x) = x^4 + 3x^3 + 7x^2 - 6x - 13, you can use the Descartes' Rule of Signs.

According to the rule, you count the number of sign changes in the coefficients of the polynomial.

First, let's list the coefficients in order:
1, 3, 7, -6, -13

We have one sign change from positive to negative (between 7 and -6), so there could be one positive real zero.

Now, let's list the coefficients in reverse order:
-13, -6, 7, 3, 1

Again, we have one sign change from negative to positive (between -6 and 7), so there could be one negative real zero.

Therefore, the number of possible positive real zeros is either 0 or an even number (in this case, 2 or 4), and the number of possible negative real zeros is also 0 or an even number (in this case, 2 or 4).

Since we have accounted for all the coefficients, any remaining zeros would be imaginary.

Hence, the possible number of imaginary zeros of g(x) = x^4 + 3x^3 + 7x^2 - 6x - 13 is 2, 4, or 0.