Write as a Set of Linear Factors f(x)=x^4-3x^3-33x^2-63x-34

$f (x) = x^{4} - 3 x^{3} - 33 x^{2} - 63 x - 34$

Factor $x^{4} - 3 x^{3} - 33 x^{2} - 63 x - 34$ 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.

$f (x) = p = \pm 1, \pm 34, \pm 2, \pm 17$

$q = \pm 1$

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

$f (x) = \pm 1, \pm 34, \pm 2, \pm 17$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

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Substitute $- 1$ into the polynomial.

$f (x) = {(- 1)}^{4} - 3 {(- 1)}^{3} - 33 {(- 1)}^{2} - 63 \cdot - 1 - 34$

Raise $- 1$ to the power of $4$ .

$f (x) = 1 - 3 {(- 1)}^{3} - 33 {(- 1)}^{2} - 63 \cdot - 1 - 34$

Raise $- 1$ to the power of $3$ .

$f (x) = 1 - 3 \cdot - 1 - 33 {(- 1)}^{2} - 63 \cdot - 1 - 34$

Multiply $- 3$ by $- 1$ .

$f (x) = 1 + 3 - 33 {(- 1)}^{2} - 63 \cdot - 1 - 34$

Add $1$ and $3$ .

$f (x) = 4 - 33 {(- 1)}^{2} - 63 \cdot - 1 - 34$

Raise $- 1$ to the power of $2$ .

$f (x) = 4 - 33 \cdot 1 - 63 \cdot - 1 - 34$

Multiply $- 33$ by $1$ .

$f (x) = 4 - 33 - 63 \cdot - 1 - 34$

Subtract $33$ from $4$ .

$f (x) = - 29 - 63 \cdot - 1 - 34$

Multiply $- 63$ by $- 1$ .

$f (x) = - 29 + 63 - 34$

Add $- 29$ and $63$ .

$f (x) = 34 - 34$

Subtract $34$ from $34$ .

$f (x) = 0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$f (x) = \frac{x^{4} - 3 x^{3} - 33 x^{2} - 63 x - 34}{x + 1}$

Divide $x^{4} - 3 x^{3} - 33 x^{2} - 63 x - 34$ by $x + 1$ .

$f (x) = x^{3} - 4 x^{2} - 29 x - 34$

Write $x^{4} - 3 x^{3} - 33 x^{2} - 63 x - 34$ as a set of factors.

$f (x) = (x + 1) (x^{3} - 4 x^{2} - 29 x - 34)$

Factor $x^{3} - 4 x^{2} - 29 x - 34$ using the rational roots test.

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Factor $x^{3} - 4 x^{2} - 29 x - 34$ 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.

$f (x) = p = \pm 1, \pm 34, \pm 2, \pm 17$

$q = \pm 1$

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

$f (x) = \pm 1, \pm 34, \pm 2, \pm 17$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

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Substitute $- 2$ into the polynomial.

$f (x) = {(- 2)}^{3} - 4 {(- 2)}^{2} - 29 \cdot - 2 - 34$

Raise $- 2$ to the power of $3$ .

$f (x) = - 8 - 4 {(- 2)}^{2} - 29 \cdot - 2 - 34$

Raise $- 2$ to the power of $2$ .

$f (x) = - 8 - 4 \cdot 4 - 29 \cdot - 2 - 34$

Multiply $- 4$ by $4$ .

$f (x) = - 8 - 16 - 29 \cdot - 2 - 34$

Subtract $16$ from $- 8$ .

$f (x) = - 24 - 29 \cdot - 2 - 34$

Multiply $- 29$ by $- 2$ .

$f (x) = - 24 + 58 - 34$

Add $- 24$ and $58$ .

$f (x) = 34 - 34$

Subtract $34$ from $34$ .

$f (x) = 0$

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$f (x) = \frac{x^{3} - 4 x^{2} - 29 x - 34}{x + 2}$

Divide $x^{3} - 4 x^{2} - 29 x - 34$ by $x + 2$ .

$f (x) = x^{2} - 6 x - 17$

Write $x^{3} - 4 x^{2} - 29 x - 34$ as a set of factors.

$f (x) = (x + 1) ((x + 2) (x^{2} - 6 x - 17))$

Remove unnecessary parentheses.

$f (x) = (x + 1) (x + 2) (x^{2} - 6 x - 17)$

Write as a Set of Linear Factors f(x)=x^4-3x^3-33x^2-63x-34

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