Factor f(x)=3x^3-16x^2-103x+36

$f (x) = 3 x^{3} - 16 x^{2} - 103 x + 36$

Factor $3 x^{3} - 16 x^{2} - 103 x + 36$ using the rational roots test.

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If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 36, \pm 2, \pm 18, \pm 3, \pm 12, \pm 4, \pm 9, \pm 6 q = \pm 1, \pm 3$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.\overline{3}, \pm 36, \pm 12, \pm 2, \pm 0.\overline{6}, \pm 18, \pm 6, \pm 3, \pm 4, \pm 1.\overline{3}, \pm 9$

Substitute $0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $0.\overline{3}$ is a root of the polynomial.

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Substitute $0.\overline{3}$ into the polynomial.

$3 \cdot {0.\overline{3}}^{3} - 16 \cdot {0.\overline{3}}^{2} - 103 \cdot 0.\overline{3} + 36$

Raise $0.\overline{3}$ to the power of $3$ .

$3 \cdot 0.\overline{037} - 16 \cdot {0.\overline{3}}^{2} - 103 \cdot 0.\overline{3} + 36$

Multiply $3$ by $0.\overline{037}$ .

$0.\overline{1} - 16 \cdot {0.\overline{3}}^{2} - 103 \cdot 0.\overline{3} + 36$

Raise $0.\overline{3}$ to the power of $2$ .

$0.\overline{1} - 16 \cdot 0.\overline{1} - 103 \cdot 0.\overline{3} + 36$

Multiply $- 16$ by $0.\overline{1}$ .

$0.\overline{1} - 1.\overline{7} - 103 \cdot 0.\overline{3} + 36$

Subtract $1.\overline{7}$ from $0.\overline{1}$ .

$- 1.\overline{6} - 103 \cdot 0.\overline{3} + 36$

Multiply $- 103$ by $0.\overline{3}$ .

$- 1.\overline{6} - 34.\overline{3} + 36$

Subtract $34.\overline{3}$ from $- 1.\overline{6}$ .

$- 36 + 36$

Add $- 36$ and $36$ .

$0$

Since $0.\overline{3}$ is a known root, divide the polynomial by $3 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{3 x^{3} - 16 x^{2} - 103 x + 36}{3 x - 1}$

Divide $3 x^{3} - 16 x^{2} - 103 x + 36$ by $3 x - 1$ .

$x^{2} - 5 x - 36$

Write $3 x^{3} - 16 x^{2} - 103 x + 36$ as a set of factors.

$(3 x - 1) (x^{2} - 5 x - 36)$

Factor $x^{2} - 5 x - 36$ using the AC method.

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Factor $x^{2} - 5 x - 36$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $- 5$ .

$- 9, 4$

Write the factored form using these integers.

$(3 x - 1) ((x - 9) (x + 4))$

Remove unnecessary parentheses.

$(3 x - 1) (x - 9) (x + 4)$

$f (x) = (3 x - 1) (x - 9) (x + 4)$

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