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

Ask by Sherman Riley. in the United States
Mar 17,2025

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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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Did you know that the Rational Root Theorem can be your best friend when searching for rational zeros? According to this theorem, if \( p/q \) is a rational zero of a polynomial, then \( p \) is a factor of the constant term (in this case, -7), and \( q \) is a factor of the leading coefficient (3). This gives you potential candidates like \( \pm 1, \pm 7, \pm \frac{1}{3}, \pm \frac{7}{3} \) to test in your function! Once you've identified these candidates, simply plug them into the polynomial function. A common mistake is neglecting to check all possibilities or forgetting to verify that the result is indeed zero. Remember, even if one candidate works, you may find more rational zeros, so keep trying each option until you’ve exhausted your choices!

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