Find all the zeros. Write the answer in exact form. $\newline$ $m(x)=5 x^{3}-x^{2}-35 x+7$ $\newline$ Part: $0$ / $6$ $\newline$ Part $1$ of $6$ $\newline$ We begin by first looking for rational zeros. We can apply the rational zero theorem because the polynomial has integer coefficients. $\newline$ $m(x)=5 x^{3}-x^{2}-35 x+7$ $\newline$ Possible rational zeros: $\newline$ Factors of $\newline$ Factors of

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Q. Find all the zeros. Write the answer in exact form.

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m(x)=5 x^{3}-x^{2}-35 x+7

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Part:

0

6

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Part

1

6

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We begin by first looking for rational zeros. We can apply the rational zero theorem because the polynomial has integer coefficients.

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m(x)=5 x^{3}-x^{2}-35 x+7

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Possible rational zeros:

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Factors of

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Factors of

Use Rational Root Theorem: To find the rational zeros of the polynomial $m(x) = 5x^3 - x^2 - 35x + 7$ , we can use the Rational Root Theorem. This theorem states that any rational zero, expressed as a fraction $\frac{p}{q}$ , will have $p$ as a factor of the constant term (in this case, $7$ ) and $q$ as a factor of the leading coefficient (in this case, $5$ ).
Test Possible Zeros: The factors of the constant term $7$ are $\pm1$ , $\pm7$ . The factors of the leading coefficient $5$ are $\pm1$ , $\pm5$ . Therefore, the possible rational zeros are $\pm1$ , $\pm7$ , $\pm\frac{1}{5}$ , and $\pm\frac{7}{5}$ .
Test $x = 1$ : We will now test these possible rational zeros by using synthetic division or direct substitution to see if they yield a zero remainder, indicating that they are indeed zeros of the polynomial.
Test $x = -1$ : Let's first test $x = 1$ . Substituting $1$ into the polynomial gives us: $\newline$ $m(1) = 5(1)^3 - (1)^2 - 35(1) + 7 = 5 - 1 - 35 + 7 = -24$ . $\newline$ Since $m(1) \neq 0$ , $x = 1$ is not a zero of the polynomial.
Test $x = 7$ : Next, let's test $x = -1$ . Substituting $-1$ into the polynomial gives us: $m(-1) = 5(-1)^3 - (-1)^2 - 35(-1) + 7 = -5 - 1 + 35 + 7 = 36$ . Since $m(-1) \neq 0$ , $x = -1$ is not a zero of the polynomial.
Test $x = -7$ : Now, let's test $x = 7$ . Substituting $7$ into the polynomial gives us: $m(7) = 5(7)^3 - (7)^2 - 35(7) + 7 = 5(343) - 49 - 245 + 7 = 1715 - 49 - 245 + 7 = 1428$ . Since $m(7) \neq 0$ , $x = 7$ is not a zero of the polynomial.
Test $x = \frac{1}{5}$ : Let's test $x = -7$ . Substituting $-7$ into the polynomial gives us: $m(-7) = 5(-7)^3 - (-7)^2 - 35(-7) + 7 = -5(343) - 49 + 245 + 7 = -1715 - 49 + 245 + 7 = -1512.$ Since $m(-7) \neq 0$ , $x = -7$ is not a zero of the polynomial.
Test $x = -\frac{1}{5}$ : Now, let's test $x = \frac{1}{5}$ . Substituting $\frac{1}{5}$ into the polynomial gives us: $\newline$ $m(\frac{1}{5}) = 5(\frac{1}{5})^3 - (\frac{1}{5})^2 - 35(\frac{1}{5}) + 7 = 5(\frac{1}{125}) - (\frac{1}{25}) - 7 + 7 = \frac{1}{25} - \frac{1}{25} = 0$ . $\newline$ Since $m(\frac{1}{5}) = 0$ , $x = \frac{1}{5}$ is a zero of the polynomial.
Find Other Zeros: Finally, let's test $x = -\frac{1}{5}$ . Substituting $-\frac{1}{5}$ into the polynomial gives us: $\newline$ $m(-\frac{1}{5}) = 5(-\frac{1}{5})^3 - (-\frac{1}{5})^2 - 35(-\frac{1}{5}) + 7 = -5(\frac{1}{125}) - (\frac{1}{25}) + 7 - 7 = -\frac{1}{25} - \frac{1}{25} = -\frac{2}{25}$ . $\newline$ Since $m(-\frac{1}{5}) \neq 0$ , $x = -\frac{1}{5}$ is not a zero of the polynomial.
Perform Synthetic Division: Now that we have found one rational zero, $x = \frac{1}{5}$ , we can use synthetic division to divide the polynomial by $(5x - 1)$ to find the other zeros.
Quotient and Simplification: Performing synthetic division with the zero $\frac{1}{5}$ , we divide the polynomial by $(5x - 1)$ . The synthetic division should look like this: $\newline$ $\frac{1}{5}$ | $5$ $-1$ $-35$ $7$ $\newline$ | $1$ $0$ $-7$ $\newline$ ---------------- $\newline$ $5$ $0$ $-35$ $0$ $\newline$ The quotient we get is $(5x - 1)$ $4$ , which simplifies to $(5x - 1)$ $5$ .
Factor Quadratic: The quadratic $x^2 - 7$ can be factored as $(x - \sqrt{7})(x + \sqrt{7})$ . Therefore, the remaining zeros of the polynomial are $x = \sqrt{7}$ and $x = -\sqrt{7}$ .