Sketch the graph of the polynomial function. Use synthetic division and the remainder theorem to find the zeros. \( f(x)=x^{3}-43 x+42 \)

To find the zeros of the polynomial function \( f(x) = x^3 - 43x + 42 \), let’s try the Rational Root Theorem. Testing potential factors, we can plug in different values of \( x \). After testing \( x = 1 \): \[ f(1) = 1^3 - 43(1) + 42 = 1 - 43 + 42 = 0. \] Since \( f(1) = 0 \), we have found a zero at \( x = 1 \). Now we can use synthetic division to divide \( f(x) \) by \( x - 1 \): \[ \begin{array}{r|rrr} 1 & 1 & 0 & -43 & 42 \\ & & 1 & 1 & -42 \\ \hline & 1 & 1 & -42 & 0 \\ \end{array} \] The result is \( x^2 + x - 42 \). Now we can factor this quadratic: \[ x^2 + x - 42 = (x - 6)(x + 7). \] Thus, the zeros of the polynomial function are \( x = 1\), \( x = 6\), and \( x = -7\). The graph of the polynomial will show zeros at these points, typically crossing the x-axis at each one. The general shape would demonstrate a cubic function, increasing from left to right with the curve touching the x-axis at these intercepts. To summarize: the polynomial has three real zeros \( x = 1\), \( x = 6\), and \( x = -7\).