Algebra Examples

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

Step 1

Factor

$x$ out of

$x^{3} - 4 x^{2} + 3 x$ .

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Step 1.1

Factor

$x$ out of

$x^{3}$ .

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

Step 1.2

Factor

$x$ out of

$- 4 x^{2}$ .

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

Step 1.3

Factor

$x$ out of

$3 x$ .

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

Step 1.4

Factor

$x$ out of

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

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

Step 1.5

Factor

$x$ out of

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

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

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

Step 2

Factor.

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Step 2.1

Factor

$x^{2} - 4 x + 3$ using the AC method.

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Step 2.1.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$3$ and whose sum is

$- 4$ .

$- 3, - 1$

Step 2.1.2

Write the factored form using these integers.

$x ((x - 3) (x - 1)) = 0$

$x ((x - 3) (x - 1)) = 0$

Step 2.2

Remove unnecessary parentheses.

$x (x - 3) (x - 1) = 0$

$x (x - 3) (x - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x = 0$

$x - 3 = 0$

$x - 1 = 0$

Step 4

Set

$x$ equal to

$0$ .

$x = 0$

Step 5

Set

$x - 3$ equal to

$0$ and solve for

$x$ .

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Step 5.1

Set

$x - 3$ equal to

$0$ .

$x - 3 = 0$

Step 5.2

Add

$3$ to both sides of the equation.

$x = 3$

$x = 3$

Step 6

Set

$x - 1$ equal to

$0$ and solve for

$x$ .

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Step 6.1

Set

$x - 1$ equal to

$0$ .

$x - 1 = 0$

Step 6.2

Add

$1$ to both sides of the equation.

$x = 1$

$x = 1$

Step 7

The final solution is all the values that make

$x (x - 3) (x - 1) = 0$ true.

$x = 0, 3, 1$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$