Write as a Set of Linear Factors x^3-9x^2+22x-12

$x^{3} - 9 x^{2} + 22 x - 12$

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.

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

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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

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Substitute $3$ into the polynomial.

$3^{3} - 9 \cdot 3^{2} + 22 \cdot 3 - 12$

Simplify each term.

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

$27 - 9 \cdot 3^{2} + 22 \cdot 3 - 12$

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

$27 - 9 \cdot 9 + 22 \cdot 3 - 12$

Multiply $- 9$ by $9$ .

$27 - 81 + 22 \cdot 3 - 12$

Multiply $22$ by $3$ .

$27 - 81 + 66 - 12$

Subtract $81$ from $27$ .

$- 54 + 66 - 12$

Add $- 54$ and $66$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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

$\frac{x^{3} - 9 x^{2} + 22 x - 12}{x - 3}$

Divide $x^{3} - 9 x^{2} + 22 x - 12$ by $x - 3$ .

$x^{2} - 6 x + 4$

Write $x^{3} - 9 x^{2} + 22 x - 12$ as a set of factors.

$(x - 3) (x^{2} - 6 x + 4)$

Write as a Set of Linear Factors x^3-9x^2+22x-12

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