Algebra Examples

$5 x {(x - 7)}^{2} (x^{2} - 16)$

Step 1

Set $5 x {(x - 7)}^{2} (x^{2} - 16)$ equal to $0$ .

$5 x {(x - 7)}^{2} (x^{2} - 16) = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor.

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Step 2.1.1

Rewrite $16$ as $4^{2}$ .

$5 x {(x - 7)}^{2} (x^{2} - 4^{2}) = 0$

Step 2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$5 x {(x - 7)}^{2} ((x + 4) (x - 4)) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

${(x - 7)}^{2} = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set ${(x - 7)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set ${(x - 7)}^{2}$ equal to $0$ .

${(x - 7)}^{2} = 0$

Step 2.4.2

Solve ${(x - 7)}^{2} = 0$ for $x$ .

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Step 2.4.2.1

Set the $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.4.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.7

The final solution is all the values that make $5 x {(x - 7)}^{2} (x^{2} - 16) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $2$ )

$x = - 4$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

$x = 0$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $2$ )

$x = - 4$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

Step 3