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.
Similar Questions
- Let \(p(x)\) be a polynomial. When \(p(x)\) is divided by \((x-1)\), it leaves 2 as the remainder. When \(p(x)\) is divided by \((x-2)\), it leaves 1 as the remainder. What is the remainder when \(p(x)\) is divided by \((x - 1)(x-2)\)?
- What is the remainder when \(x^6\) is divided by \(x^2 + 1\)?
- \((x+2)\) is a factor of which one of the following?
What is the HCF of the polynomials x⁸ + x⁴ + 1 and x⁴ + x² + 1?
- Suppose \(p(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0\) and \(q(x) = x^4 + b_3x^3 + b_2x^2 + b_1x + b_0\) are the polynomials. If \(\alpha, \beta, \gamma, \delta\) are zeros of \(p(x)\) and \(\alpha, \beta, \gamma, \lambda\) are zeros of \(q(x)\), then what is \(\frac{p(x) - q(x)}{(x - \alpha) (x - \beta) (x - \gamma)}\) equal to ?
- Consider the following in respect of the polynomial \(x^{4k} + x^{4k+2} + x^{4k+4} + x^{4k+6}\):
1. The remainder is zero when the polynomial is divided by \(x^2 + 1\).
2. The remainder is zero when the polynomial is divided by \(x^4 + 1\).
Which of the statements given above is/are correct? - If \(x^3 + px^2 + qx + r\) is an integer for all integral values of \(x\), then consider the following statements :
I. \(p\) must be an integer
II. \(q\) must be an integer
III. \(r\) must be an integer
Which of the statements given above is/are correct? - Which of the following expressions can divide both the polynomials \(x^3+2x^2-5x+2\) and \(x^3+4x^2+x-6\) exactly ?
I. \(x-1\)
II. \(x+1\)
III. \(x+2\)
Select the correct answer using the code given below : If x² - 5x + 4 is a factor of x⁴ - px² + q, then what are the values of p and q respectively?
Important Questions from Polynomials
If y 2= y + 7, then what is the value of y 3?
Factorize x 2- y 2- 9z 2+ 6yz
If one of the zeros of the polynomial x 3+ ax 2+ bx + c is - 1, then the product of other two zeros is equal to :
If a(a + b + c) 2 = 1792; b(a + b + c) 2 = 1536; c(a + b + c) 2 = 768, then what will be the value of b?
If x = 3 so, what is the value of x 2 + 2x + 5 ?
