Questions: For the polynomial P(x)=3x^3+8x^2-4x-3, -3 is one zero of P. Find all the other zeros analytically. The other zeros of P(x) are.

Transcript text: For the polynomial $P(x)=3 x^{3}+8 x^{2}-4 x-3,-3$ is one zero of $P$. Find all the other zeros analytically. The other zeros of $\mathrm{P}(\mathrm{x})$ are $\square$.

Solution

Solution Steps

To find the zeros of the polynomial $ P(x) = 3x^3 + 8x^2 - 4x - 3 $ given that $-3$ is one of the zeros, we can use polynomial division to divide $ P(x) $ by $ (x + 3) $. This will give us a quadratic polynomial. We can then solve the quadratic polynomial using the quadratic formula to find the remaining zeros.

Step 1: Given Polynomial and Known Zero

We are given the polynomial $ P(x) = 3x^3 + 8x^2 - 4x - 3 $ and we know that $-3$ is one of its zeros.

Step 2: Polynomial Division

To find the other zeros, we perform polynomial division of $ P(x) $ by $ (x + 3) $. This gives us a quotient which is a quadratic polynomial.

Step 3: Solving the Quadratic Polynomial

The quotient from the polynomial division is $ 3x^2 - x - 1 $. We solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 3 $, $ b = -1 $, and $ c = -1 $.

Step 4: Calculating the Zeros

Substituting the values into the quadratic formula, we get: \[ x = \frac{1 \pm \sqrt{1 + 12}}{6} = \frac{1 \pm \sqrt{13}}{6} \] Thus, the zeros are: \[ x = \frac{1 - \sqrt{13}}{6} \quad \text{and} \quad x = \frac{1 + \sqrt{13}}{6} \]