Algebra Examples

$f (x) = {(x - 3)}^{3}$

Replace $f (x)$ with $y$ .

$y = {(x - 3)}^{3}$

To find the roots of the equation, replace $y$ with $0$ and solve.

$0 = x^{3} - 9 x^{2} + 27 x - 27$

Rewrite the equation as $x^{3} - 9 x^{2} + 27 x - 27 = 0$ .

$x^{3} - 9 x^{2} + 27 x - 27 = 0$

Factor the left side of the equation.

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Factor $x^{3} - 9 x^{2} + 27 x - 27$ 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.

$p = \pm 1, \pm 3, \pm 9, \pm 27 q = \pm 1$

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

$\pm 1, \pm 3, \pm 9, \pm 27$

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.

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

Simplify each term.

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

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

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

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

Multiply $- 9$ by $9$ .

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

Multiply $27$ by $3$ .

$27 - 81 + 81 - 27$

Subtract $81$ from $27$ .

$- 54 + 81 - 27$

Add $- 54$ and $81$ .

$27 - 27$

Subtract $27$ from $27$ .

$0$

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.

$\frac{x^{3} - 9 x^{2} + 27 x - 27}{x - 3}$

Divide $x^{3} - 9 x^{2} + 27 x - 27$ by $x - 3$ .

$x^{2} - 6 x + 9$

Write $x^{3} - 9 x^{2} + 27 x - 27$ as a set of factors.

$(x - 3) (x^{2} - 6 x + 9) = 0$

Factor using the perfect square rule.

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Rewrite $9$ as $3^{2}$ .

$(x - 3) (x^{2} - 6 x + 3^{2}) = 0$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot x \cdot - 3$

Simplify.

$2 a b = - 6 x$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = - 3$ .

$(x - 3) {(x - 3)}^{2} = 0$

Combine like factors.

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

$(x - 3) {(x - 3)}^{2} = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(x - 3)}^{1 + 2} = 0$

Add $1$ and $2$ .

${(x - 3)}^{3} = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The solution is the result of $x - 3 = 0$ .

$x = 3$

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