Questions: Find all zeros. One zero has been given. 9) f(x)=3x^3+2x^2-19x+6; 2

Solution Steps

To find all zeros of the polynomial \( f(x) = 3x^3 + 2x^2 - 19x + 6 \), we can use the given zero \( x = 2 \) to perform polynomial division and reduce the polynomial to a quadratic. Then, we solve the resulting quadratic equation to find the remaining zeros.

We are given that \( x = 2 \) is a zero of the polynomial \( f(x) = 3x^3 + 2x^2 - 19x + 6 \). Using this zero, we perform synthetic division to reduce the polynomial to a quadratic. The result of the division is a quadratic polynomial: \( 3x^2 + 8x - 9 \).

Next, we solve the quadratic equation \( 3x^2 + 8x - 9 = 0 \) to find the remaining zeros. Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 8 \), and \( c = -9 \), we find the solutions to be: \[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-9)}}{2 \cdot 3} \] \[ x = \frac{-8 \pm \sqrt{64 + 108}}{6} \] \[ x = \frac{-8 \pm \sqrt{172}}{6} \] \[ x = \frac{-8 \pm 13.114}{6} \]

This gives us the solutions: \[ x_1 = \frac{-8 + 13.114}{6} \approx 0.3333 \] \[ x_2 = \frac{-8 - 13.114}{6} \approx -3.0000 \]

Final Answer