Jhe function below has at least one rational zero. \[ h(x)=3 x^{3}-10 x^{2}+10 x-7 \]

We start by using the Rational Root Theorem. The theorem tells us that any rational zero of \[ h(x)=3x^3-10x^2+10x-7 \] has the form \[ \frac{p}{q}, \] where \( p \) divides the constant term \(-7\) and \( q \) divides the leading coefficient \(3\). Thus the possible values for \( p \) are \(\pm1, \pm7\) and for \( q \) are \(\pm1, \pm3\). The potential rational zeros are therefore \[ \pm1, \quad \pm7, \quad \pm\frac{1}{3}, \quad \pm\frac{7}{3}. \] Next, we test one of these candidates. We begin with \( x = \frac{7}{3} \): \[ h\left(\frac{7}{3}\right)= 3\left(\frac{7}{3}\right)^3 - 10\left(\frac{7}{3}\right)^2 + 10\left(\frac{7}{3}\right) - 7. \] Compute each term: 1. \(\left(\frac{7}{3}\right)^3 = \frac{343}{27}\) so \(3\left(\frac{343}{27}\right)= \frac{1029}{27}\). 2. \(\left(\frac{7}{3}\right)^2 = \frac{49}{9}\) so \(10\left(\frac{49}{9}\right)= \frac{490}{9}=\frac{490\cdot3}{9\cdot3}=\frac{1470}{27}\). 3. \(10\left(\frac{7}{3}\right)= \frac{70}{3}=\frac{70\cdot9}{3\cdot9}=\frac{630}{27}\). 4. Rewrite \(-7\) as \(-\frac{189}{27}\). Now, combine all terms using the common denominator 27: \[ h\left(\frac{7}{3}\right)= \frac{1029}{27} - \frac{1470}{27} + \frac{630}{27} - \frac{189}{27} = \frac{1029 - 1470 + 630 -189}{27}. \] Simplify the numerator: \[ 1029 - 1470 = -441, \quad -441 + 630 = 189, \quad 189 - 189 = 0. \] Thus, \[ h\left(\frac{7}{3}\right)= \frac{0}{27} = 0. \] Since \( h\left(\frac{7}{3}\right)=0 \), we have found a rational zero: \[ x=\frac{7}{3}. \] To factor further, we divide \( h(x) \) by \( \left(3x-7\right) \) (since setting \( 3x-7=0 \) gives \( x=\frac{7}{3} \)). The polynomial division yields: \[ h(x)=(3x-7)(x^2-x+1). \] The quadratic \( x^2 - x + 1 \) has discriminant: \[ \Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3, \] so it does not have any real (or rational) zeros. Thus, the only rational zero of \( h(x) \) is \[ x=\frac{7}{3}. \]