Precalculus Examples

$y = 4 x^{3} - 20 x^{2} + 25 x$

Step 1

Set $4 x^{3} - 20 x^{2} + 25 x$ equal to $0$ .

$4 x^{3} - 20 x^{2} + 25 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $x$ out of $4 x^{3} - 20 x^{2} + 25 x$ .

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Step 2.1.1.1

Factor $x$ out of $4 x^{3}$ .

$x (4 x^{2}) - 20 x^{2} + 25 x = 0$

Step 2.1.1.2

Factor $x$ out of $- 20 x^{2}$ .

$x (4 x^{2}) + x (- 20 x) + 25 x = 0$

Step 2.1.1.3

Factor $x$ out of $25 x$ .

$x (4 x^{2}) + x (- 20 x) + x \cdot 25 = 0$

Step 2.1.1.4

Factor $x$ out of $x (4 x^{2}) + x (- 20 x)$ .

$x (4 x^{2} - 20 x) + x \cdot 25 = 0$

Step 2.1.1.5

Factor $x$ out of $x (4 x^{2} - 20 x) + x \cdot 25$ .

$x (4 x^{2} - 20 x + 25) = 0$

Step 2.1.2

Factor using the perfect square rule.

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Step 2.1.2.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$x ({(2 x)}^{2} - 20 x + 25) = 0$

Step 2.1.2.2

Rewrite $25$ as $5^{2}$ .

$x ({(2 x)}^{2} - 20 x + 5^{2}) = 0$

Step 2.1.2.3

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

$20 x = 2 \cdot (2 x) \cdot 5$

Step 2.1.2.4

Rewrite the polynomial.

$x ({(2 x)}^{2} - 2 \cdot (2 x) \cdot 5 + 5^{2}) = 0$

Step 2.1.2.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 5$ .

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

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 $0$ .

$x = 0$

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

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set ${(2 x - 5)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

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

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

Step 2.4.2

Solve ${(2 x - 5)}^{2} = 0$ for $x$ .

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Step 2.4.2.1

Set the $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 2.4.2.2

Solve for $x$ .

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Step 2.4.2.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 2.4.2.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 2.4.2.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 2.4.2.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 2.4.2.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 2.5

The final solution is all the values that make $x {(2 x - 5)}^{2} = 0$ true.

$x = 0, \frac{5}{2}$

Step 3