Understanding the Problem

The question asks us to find the product of the other two zeros of a cubic polynomial \(P(x) = x^3 + ax^2 + bx + c\), given that one of its zeros is -1.

Properties of Polynomial Zeros

For a general cubic polynomial of the form \(Ax^3 + Bx^2 + Cx + D\), with zeros (roots) denoted by \(\alpha\), \(\beta\), and \(\gamma\), the following relationships hold:

• Sum of the zeros: \(\alpha + \beta + \gamma = -B/A\)
• Sum of the products of the zeros taken two at a time: \(\alpha\beta + \beta\gamma + \gamma\alpha = C/A\)
• Product of the zeros: \(\alpha\beta\gamma = -D/A\)

These relationships are often referred to as Vieta's formulas.

Applying Properties to the Given Polynomial

The given polynomial is \(x^3 + ax^2 + bx + c\). Comparing this with the general form \(Ax^3 + Bx^2 + Cx + D\), we have:

• \(A = 1\)
• \(B = a\)
• \(C = b\)
• \(D = c\)

Let the three zeros of the polynomial be \(\alpha\), \(\beta\), and \(\gamma\). We are given that one zero is -1. Let's assume \(\alpha = -1\).

Now, let's use Vieta's formulas with the given polynomial and the known zero \(\alpha = -1\):

1. Sum of the zeros:

\(\alpha + \beta + \gamma = -B/A\)

\(-1 + \beta + \gamma = -a/1\)

\(-1 + \beta + \gamma = -a\)

Rearranging this gives us the sum of the other two zeros:

\(\beta + \gamma = -a + 1\)

2. Sum of the products of the zeros taken two at a time:

\(\alpha\beta + \beta\gamma + \gamma\alpha = C/A\)

Substitute \(\alpha = -1\):

\((-1)\beta + \beta\gamma + \gamma(-1) = b/1\)

\(-\beta + \beta\gamma - \gamma = b\)

Rearrange the terms:

\(\beta\gamma - (\beta + \gamma) = b\)

3. Product of the zeros:

\(\alpha\beta\gamma = -D/A\)

Substitute \(\alpha = -1\):

\((-1)\beta\gamma = -c/1\)

\(-\beta\gamma = -c\)

Multiplying by -1 gives us the product of the other two zeros directly:

\(\beta\gamma = c\)

Finding the Product of the Other Two Zeros

We need to find the value of \(\beta\gamma\). From the relationship for the sum of products taken two at a time, we have:

\(\beta\gamma - (\beta + \gamma) = b\)

We previously found that \(\beta + \gamma = -a + 1\). Substitute this into the equation:

\(\beta\gamma - (-a + 1) = b\)

\(\beta\gamma + a - 1 = b\)

Now, solve for \(\beta\gamma\):

\(\beta\gamma = b - a + 1\)

This result matches the value obtained from the product of zeros formula only if c = b - a + 1. Let's check our logic.

The fact that -1 is a zero means that \(P(-1) = 0\). Let's substitute \(x = -1\) into the polynomial:

\((-1)^3 + a(-1)^2 + b(-1) + c = 0\)

\(-1 + a(1) - b + c = 0\)

\(-1 + a - b + c = 0\)

From this equation, we can express c in terms of a and b:

\(c = 1 - a + b\)

\(c = b - a + 1\)

So, the product of the three zeros is \(\alpha\beta\gamma = (-1)\beta\gamma\). We also know this is equal to -c.

\((-1)\beta\gamma = -c\)

\(\beta\gamma = c\)

Substitute the value of c we found from \(P(-1) = 0\):

\(\beta\gamma = (b - a + 1)\)

Therefore, the product of the other two zeros (\(\beta\) and \(\gamma\)) is \(b - a + 1\).

Comparing with Options

Let's compare our result with the given options:

1. \(b - a + 1\)
2. \(b - a - 1\)
3. \(a - b + 1\)
4. \(a - b - 1\)

Our calculated product of the other two zeros is \(b - a + 1\), which matches Option 1.

Conclusion

Given that -1 is a zero of the polynomial \(x^3 + ax^2 + bx + c\), substituting \(x=-1\) into the polynomial equation \(P(x)=0\) gives the condition relating the coefficients a, b, and c. Using Vieta's formulas for the product of zeros, and substituting the known zero and the relation between coefficients, we found the product of the other two zeros.

The product of the other two zeros is \(b - a + 1\).