Answer
The function \( h(x) = 3x^3 - 10x^2 + 10x - 7 \) has a rational zero at \( x = \frac{7}{3} \).
Solution
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
\[
x=\frac{7}{3}.
\]
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