Find the x and y Intercepts f(x)=x^4-2x^3+x^2

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

Find the x-intercepts.

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

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

Solve the equation.

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

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

Factor the left side of the equation.

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

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

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

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

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

Multiply by $1$ .

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

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

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

Factor $x^{2}$ out of $x^{2} (x^{2} - 2 x) + x^{2} \cdot 1$ .

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

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

$x^{2} = 0$

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

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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

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

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $x^{2} {(x - 1)}^{2} = 0$ true.

$x = 0, 1$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (1, 0)$

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

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

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

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

Multiply $- 2$ by $0$ .

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

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

$y = 0 + 0 + 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

x-intercept(s): $(0, 0), (1, 0)$

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

Find the x and y Intercepts f(x)=x^4-2x^3+x^2

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