Finite Math Examples

$y^{3} - 12 y^{2} + 35 y = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $y$ out of $y^{3} - 12 y^{2} + 35 y$ .

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Step 1.1.1

Factor $y$ out of $y^{3}$ .

$y \cdot y^{2} - 12 y^{2} + 35 y = 0$

Step 1.1.2

Factor $y$ out of $- 12 y^{2}$ .

$y \cdot y^{2} + y (- 12 y) + 35 y = 0$

Step 1.1.3

Factor $y$ out of $35 y$ .

$y \cdot y^{2} + y (- 12 y) + y \cdot 35 = 0$

Step 1.1.4

Factor $y$ out of $y \cdot y^{2} + y (- 12 y)$ .

$y (y^{2} - 12 y) + y \cdot 35 = 0$

Step 1.1.5

Factor $y$ out of $y (y^{2} - 12 y) + y \cdot 35$ .

$y (y^{2} - 12 y + 35) = 0$

Step 1.2

Factor.

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Step 1.2.1

Factor $y^{2} - 12 y + 35$ using the AC method.

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Step 1.2.1.1

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 $35$ and whose sum is $- 12$ .

$- 7, - 5$

Step 1.2.1.2

Write the factored form using these integers.

$y ((y - 7) (y - 5)) = 0$

Step 1.2.2

Remove unnecessary parentheses.

$y (y - 7) (y - 5) = 0$

Step 2

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

$y = 0$

$y - 7 = 0$

$y - 5 = 0$

Step 3

Set $y$ equal to $0$ .

$y = 0$

Step 4

Set $y - 7$ equal to $0$ and solve for $y$ .

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Step 4.1

Set $y - 7$ equal to $0$ .

$y - 7 = 0$

Step 4.2

Add $7$ to both sides of the equation.

$y = 7$

Step 5

Set $y - 5$ equal to $0$ and solve for $y$ .

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Step 5.1

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Step 5.2

Add $5$ to both sides of the equation.

$y = 5$

Step 6

The final solution is all the values that make $y (y - 7) (y - 5) = 0$ true.

$y = 0, 7, 5$