Question

1. $f(x) = 3x^3 - 7x^2 - 13x + 17$ with a zero of $x = 1$

Explanation:

Step1: Factor the cubic polynomial

Since \( x = 1 \) is a zero of \( f(x)=3x^{3}-7x^{2}-13x + 17 \), we know that \( (x - 1) \) is a factor. We can use polynomial long division or synthetic division. From the synthetic division (as shown in the image), we get the quotient \( 3x^{2}-4x - 17 \) when we divide \( f(x) \) by \( (x - 1) \). So, \( f(x)=(x - 1)(3x^{2}-4x - 17) \).

Step2: Find the other zeros

To find the other zeros, we solve the quadratic equation \( 3x^{2}-4x - 17=0 \) using the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 3 \), \( b=-4 \), and \( c=-17 \).

First, calculate the discriminant \( \Delta=b^{2}-4ac=(-4)^{2}-4\times3\times(-17)=16 + 204 = 220 \).

Then, \( x=\frac{4\pm\sqrt{220}}{6}=\frac{4\pm2\sqrt{55}}{6}=\frac{2\pm\sqrt{55}}{3} \).

So the zeros of the function \( f(x) \) are \( x = 1 \), \( x=\frac{2+\sqrt{55}}{3} \), and \( x=\frac{2-\sqrt{55}}{3} \).

Answer:

The zeros of \( f(x)=3x^{3}-7x^{2}-13x + 17 \) are \( \boldsymbol{x = 1} \), \( \boldsymbol{x=\frac{2+\sqrt{55}}{3}} \), and \( \boldsymbol{x=\frac{2-\sqrt{55}}{3}} \)