Pre-Algebra Examples

$(9 x^{2} - 24 x + 16) = 0$

Step 1

Factor using the perfect square rule.

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Step 1.1

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} - 24 x + 16 = 0$

Step 1.2

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

${(3 x)}^{2} - 24 x + 4^{2} = 0$

Step 1.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$24 x = 2 \cdot (3 x) \cdot 4$

Step 1.4

Rewrite the polynomial.

${(3 x)}^{2} - 2 \cdot (3 x) \cdot 4 + 4^{2} = 0$

Step 1.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 4$ .

${(3 x - 4)}^{2} = 0$

Step 2

Set the $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Step 3

Solve for $x$ .

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Step 3.1

Add $4$ to both sides of the equation.

$3 x = 4$

Step 3.2

Divide each term in $3 x = 4$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{4}{3}$