the equation mx^3+mx^2+mx+c=0 has a root at x=c. given that m and c are both real and m,c does not = 0, determine the possible values of m


Given the equation:

\[
mx^3 + mx^2 + mx + c = 0
\]

and knowing that \( x = c \) is a root of the equation, we want to determine the possible values of \( m \), given that \( m \) and \( c \) are both real numbers and neither is zero.

### Step 1: Substitute \( x = c \) into the equation

Since \( x = c \) is a root, it satisfies the equation. Substitute \( x = c \):

\[
m(c^3) + m(c^2) + m(c) + c = 0
\]

Simplify this:

\[
mc^3 + mc^2 + mc + c = 0
\]

Factor \( c \) from the terms on the left side:

\[
c(mc^2 + mc + m + 1) = 0
\]

### Step 2: Analyze the equation

Since \( c \neq 0 \), divide both sides by \( c \):

\[
mc^2 + mc + m + 1 = 0
\]

### Step 3: Factorize and simplify

We can rewrite the equation by factoring out \( m \):

\[
m(c^2 + c + 1) + 1 = 0
\]

Solving for \( m \):

\[
m(c^2 + c + 1) = -1
\]

\[
m = \frac{-1}{c^2 + c + 1}
\]

### Step 4: Determine the restrictions on \( m \)

The denominator \( c^2 + c + 1 \) must not be zero since \( m \) is finite. Check if \( c^2 + c + 1 = 0 \) has real roots:

\[
c = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}
\]

The roots are complex, so \( c^2 + c + 1 \) is never zero for real \( c \). Therefore, the expression \( m = \frac{-1}{c^2 + c + 1} \) is always defined for real \( c \).

### Conclusion

The possible values of \( m \) are:

\[
m = \frac{-1}{c^2 + c + 1}, \quad \text{where } c \text{ is any real number except zero}.
\]

Would you like further details or have any questions?

### Related Questions:

1. How does the value of \( m \) change if \( c \) is positive or negative?
2. What are the implications of \( m \) being negative given the above formula?
3. How would the value of \( m \) be affected if the root were \( x = -c \) instead of \( x = c \)?
4. Can this equation have any repeated roots other than \( x = c \)?
5. How can the discriminant of a cubic equation help in analyzing the nature of its roots?

**Tip:** Always check if the denominator in a fraction involving variables can ever be zero to avoid undefined values in the solution.

The equation mx^3 + mx^2 + mx + c = 0 has a root at x = c. Given that m and c are both real and m, c ≠ 0, determine the possible values of m.

Explore how to determine the possible values of m in the cubic equation mx^3 + mx^2 + mx + c = 0 when x = c is a root. This solution uses key algebraic techniques including substitution and factoring, making it suitable for students in Grades 11-12.

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