Understanding the Polynomial and Its Zeros

We are given a cubic polynomial: \(p(x) = x^3 + 3x^2 - 6x - a\) We are also told that 2 is a zero of this polynomial. This means that when we substitute \(x=2\) into the polynomial, the result should be 0.

Step 1: Finding the Value of 'a'

Since 2 is a zero, we have \(p(2) = 0\). Let's substitute \(x=2\) into the polynomial equation:

\(p(2) = (2)^3 + 3(2)^2 - 6(2) - a = 0\)

Now, we calculate the values:

• \((2)^3 = 8\)
• \(3(2)^2 = 3(4) = 12\)
• \(6(2) = 12\)

Substituting these back into the equation:

\(8 + 12 - 12 - a = 0\)

Simplifying the equation:

\(8 - a = 0\)

Solving for \(a\):

\(a = 8\)

So, the complete polynomial is \(p(x) = x^3 + 3x^2 - 6x - 8\).

Step 2: Using Vieta's Formulas

Let the zeros of the cubic polynomial \(p(x) = Ax^3 + Bx^2 + Cx + D\) be \(\alpha\), \(\beta\), and \(\gamma\). According to Vieta's formulas, we have the following relationships:

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

For our polynomial \(p(x) = x^3 + 3x^2 - 6x - 8\), we have \(A=1\), \(B=3\), \(C=-6\), and \(D=-8\). We know that one of the zeros is 2. Let's set \(\alpha = 2\). We need to find the sum of the squares of the other two zeros, which are \(\beta\) and \(\gamma\). We need to calculate \(\beta^2 + \gamma^2\).

Sum of Zeros

Using the first Vieta's formula:

\(\alpha + \beta + \gamma = -B/A = -3/1 = -3\)

Since \(\alpha = 2\), we substitute this value:

\(2 + \beta + \gamma = -3\)

Now, we find the sum of the other two zeros (\(\beta + \gamma\)):

\(\beta + \gamma = -3 - 2\)

\(\beta + \gamma = -5\)

Sum of Product of Zeros Taken Two at a Time

Using the second Vieta's formula:

\(\alpha\beta + \alpha\gamma + \beta\gamma = C/A = -6/1 = -6\)

Substitute \(\alpha = 2\) into this equation:

\(2\beta + 2\gamma + \beta\gamma = -6\)

Factor out 2 from the first two terms:

\(2(\beta + \gamma) + \beta\gamma = -6\)

We already found that \(\beta + \gamma = -5\). Substitute this value:

\(2(-5) + \beta\gamma = -6\)

\(-10 + \beta\gamma = -6\)

Now, solve for the product of the other two zeros (\(\beta\gamma\)):

\(\beta\gamma = -6 + 10\)

\(\beta\gamma = 4\)

Step 3: Calculating the Sum of the Squares of the Other Zeros

We need to find \(\beta^2 + \gamma^2\). We can use the algebraic identity:

\((\beta + \gamma)^2 = \beta^2 + \gamma^2 + 2\beta\gamma\)

Rearranging this formula to solve for \(\beta^2 + \gamma^2\):

\(\beta^2 + \gamma^2 = (\beta + \gamma)^2 - 2\beta\gamma\)

We have the values for \(\beta + \gamma = -5\) and \(\beta\gamma = 4\). Substitute these values into the equation:

\(\beta^2 + \gamma^2 = (-5)^2 - 2(4)\)

Calculate the terms:

• \((-5)^2 = 25\)
• \(2(4) = 8\)

Substitute these results back:

\(\beta^2 + \gamma^2 = 25 - 8\)

\(\beta^2 + \gamma^2 = 17\)

Conclusion

The sum of the squares of the other zeros of the polynomial \(p(x) = x^3 + 3x^2 - 6x - a\), given that 2 is one of its zeros, is 17.