Verify that 3,−1,−133, -1, -\dfrac{1}{3} are the zeroes of the cubic polynomial p(x)=3x3−5x2−11x−3p(x) = 3x^3 - 5x^2 - 11x - 3, and then verify the relationship between the zeroes and the coefficients.

Answer: The values 3,−1,−133, -1, -\frac{1}{3} are verified to be the zeroes of the polynomial p(x)=3x3−5x2−11x−3p(x) = 3x^3 - 5x^2 - 11x - 3. The relationships between the zeroes and the coefficients are also verified as follows: α+β+γ=53=−ba\alpha + \beta + \gamma = \frac{5}{3} = -\frac{b}{a} αβ+βγ+γα=−113=ca\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{11}{3} = \frac{c}{a} αβγ=1=−da\alpha\beta\gamma = 1 = -\frac{d}{a}

Step-by-step solution

Step 1: Evaluate p(x)p(x) at x=3x=3

To verify that a number is a zero of a polynomial, we substitute that number into the polynomial and check if the result is zero. We will start by substituting x=3x=3 into the polynomial p(x)=3x3−5x2−11x−3p(x) = 3x^3 - 5x^2 - 11x - 3.

Step 2: Calculate p(3)p(3)

After substituting x=3x=3, we perform the arithmetic operations: 33=273^3 = 27, 32=93^2 = 9. Then, we calculate the products and sums to find the value of p(3)p(3). Since p(3)=0p(3)=0, x=3x=3 is indeed a zero of the polynomial.

Step 3: Evaluate p(x)p(x) at x=−1x=-1

Next, we substitute x=−1x=-1 into the polynomial p(x)p(x) to check if it is also a zero. We replace every xx in the expression with −1-1.

Step 4: Calculate p(−1)p(-1)

We calculate (−1)3=−1(-1)^3 = -1 and (−1)2=1(-1)^2 = 1. Then we perform the multiplications and additions. Since p(−1)=0p(-1)=0, x=−1x=-1 is also a zero of the polynomial.

Step 5: Evaluate p(x)p(x) at x=−13x=-\frac{1}{3}

Finally, we substitute x=−13x=-\frac{1}{3} into the polynomial p(x)p(x) to verify the last given zero. We carefully substitute the fractional value into the expression.

Step 6: Calculate p(−13)p(-\frac{1}{3})

We calculate (−13)3=−127(-\frac{1}{3})^3 = -\frac{1}{27} and (−13)2=19(-\frac{1}{3})^2 = \frac{1}{9}. Then we simplify the expression by finding a common denominator for the fractions. Since p(−13)=0p(-\frac{1}{3})=0, x=−13x=-\frac{1}{3} is also a zero of the polynomial. Thus, all three given values are verified zeros.

Step 7: Identify coefficients of p(x)p(x)

Now we will verify the relationship between the zeroes and the coefficients. A general cubic polynomial is given by ax3+bx2+cx+dax^3 + bx^2 + cx + d. By comparing this with p(x)=3x3−5x2−11x−3p(x) = 3x^3 - 5x^2 - 11x - 3, we can identify the coefficients a,b,c,da, b, c, d.

Step 8: Assign values to zeroes

Let the zeroes be α\alpha, β\beta, and γ\gamma. We assign the given verified zeroes to these variables for convenience in applying the relationships.

Step 9: Verify sum of zeroes

The relationship for the sum of the zeroes of a cubic polynomial is α+β+γ=−ba\alpha + \beta + \gamma = -\frac{b}{a}. We calculate the sum of our zeroes and compare it with −b/a-b/a using the coefficients identified earlier. Both values match, verifying this relationship.

Step 10: Verify sum of products of zeroes taken two at a time

The relationship for the sum of the products of the zeroes taken two at a time is αβ+βγ+γα=ca\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}. We compute this sum using our zeroes and compare it with c/ac/a. Both results are equal, confirming this relationship.

Step 11: Verify product of zeroes

The relationship for the product of the zeroes of a cubic polynomial is αβγ=−da\alpha\beta\gamma = -\frac{d}{a}. We calculate the product of our zeroes and compare it with −d/a-d/a. The values match, which verifies the final relationship.

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