Precalculus Examples

$f (x) = x^{4} - 8 x^{3} + 16 x^{2}$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = x^{4} - 8 x^{3} + 16 x^{2}$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $x^{4} - 8 x^{3} + 16 x^{2} = 0$ .

$x^{4} - 8 x^{3} + 16 x^{2} = 0$

Step 1.2.2

Factor the left side of the equation.

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Step 1.2.2.1

Factor $x^{2}$ out of $x^{4} - 8 x^{3} + 16 x^{2}$ .

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Step 1.2.2.1.1

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} - 8 x^{3} + 16 x^{2} = 0$

Step 1.2.2.1.2

Factor $x^{2}$ out of $- 8 x^{3}$ .

$x^{2} x^{2} + x^{2} (- 8 x) + 16 x^{2} = 0$

Step 1.2.2.1.3

Factor $x^{2}$ out of $16 x^{2}$ .

$x^{2} x^{2} + x^{2} (- 8 x) + x^{2} \cdot 16 = 0$

Step 1.2.2.1.4

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} (- 8 x)$ .

$x^{2} (x^{2} - 8 x) + x^{2} \cdot 16 = 0$

Step 1.2.2.1.5

Factor $x^{2}$ out of $x^{2} (x^{2} - 8 x) + x^{2} \cdot 16$ .

$x^{2} (x^{2} - 8 x + 16) = 0$

Step 1.2.2.2

Factor using the perfect square rule.

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Step 1.2.2.2.1

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

$x^{2} (x^{2} - 8 x + 4^{2}) = 0$

Step 1.2.2.2.2

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

$8 x = 2 \cdot x \cdot 4$

Step 1.2.2.2.3

Rewrite the polynomial.

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

Step 1.2.2.2.4

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

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

Step 1.2.3

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

$x^{2} = 0$

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

Step 1.2.4

Set $x^{2}$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 1.2.4.2

Solve $x^{2} = 0$ for $x$ .

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Step 1.2.4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 1.2.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 1.2.4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.2.4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 1.2.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.2.5

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 1.2.5.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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Step 1.2.5.2.1

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

$x - 4 = 0$

Step 1.2.5.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 1.2.6

The final solution is all the values that make $x^{2} {(x - 4)}^{2} = 0$ true.

$x = 0, 4$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (4, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(0)}^{4} - 8 {(0)}^{3} + 16 {(0)}^{2}$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = 0^{4} - 8 {(0)}^{3} + 16 {(0)}^{2}$

Step 2.2.2

Remove parentheses.

$y = 0^{4} - 8 \cdot 0^{3} + 16 {(0)}^{2}$

Step 2.2.3

Remove parentheses.

$y = 0^{4} - 8 \cdot 0^{3} + 16 \cdot 0^{2}$

Step 2.2.4

Remove parentheses.

$y = {(0)}^{4} - 8 {(0)}^{3} + 16 {(0)}^{2}$

Step 2.2.5

Simplify ${(0)}^{4} - 8 {(0)}^{3} + 16 {(0)}^{2}$ .

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Step 2.2.5.1

Simplify each term.

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Step 2.2.5.1.1

Raising $0$ to any positive power yields $0$ .

$y = 0 - 8 {(0)}^{3} + 16 {(0)}^{2}$

Step 2.2.5.1.2

Raising $0$ to any positive power yields $0$ .

$y = 0 - 8 \cdot 0 + 16 {(0)}^{2}$

Step 2.2.5.1.3

Multiply $- 8$ by $0$ .

$y = 0 + 0 + 16 {(0)}^{2}$

Step 2.2.5.1.4

Raising $0$ to any positive power yields $0$ .

$y = 0 + 0 + 16 \cdot 0$

Step 2.2.5.1.5

Multiply $16$ by $0$ .

$y = 0 + 0 + 0$

Step 2.2.5.2

Simplify by adding numbers.

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Step 2.2.5.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 2.2.5.2.2

Add $0$ and $0$ .

$y = 0$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, 0)$

Step 3

List the intersections.

x-intercept(s): $(0, 0), (4, 0)$

y-intercept(s): $(0, 0)$

Step 4