Basic Math Examples

$(1 - y) ((2 - y) (- 3 - y)) - 24 + 12 y = 0$

Step 1

Simplify $(1 - y) (2 - y) (- 3 - y) - 24 + 12 y$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Expand $(1 - y) (2 - y)$ using the FOIL Method.

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Step 1.1.1.1

Apply the distributive property.

$(1 (2 - y) - y (2 - y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.1.2

Apply the distributive property.

$(1 \cdot 2 + 1 (- y) - y (2 - y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.1.3

Apply the distributive property.

$(1 \cdot 2 + 1 (- y) - y \cdot 2 - y (- y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2

Simplify and combine like terms.

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Step 1.1.2.1

Simplify each term.

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Step 1.1.2.1.1

Multiply $2$ by $1$ .

$(2 + 1 (- y) - y \cdot 2 - y (- y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.2

Multiply $- y$ by $1$ .

$(2 - y - y \cdot 2 - y (- y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.3

Multiply $2$ by $- 1$ .

$(2 - y - 2 y - y (- y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.4

Rewrite using the commutative property of multiplication.

$(2 - y - 2 y - 1 \cdot - 1 y \cdot y) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 1.1.2.1.5.1

Move $y$ .

$(2 - y - 2 y - 1 \cdot - 1 (y \cdot y)) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.5.2

Multiply $y$ by $y$ .

$(2 - y - 2 y - 1 \cdot - 1 y^{2}) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.6

Multiply $- 1$ by $- 1$ .

$(2 - y - 2 y + 1 y^{2}) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.1.7

Multiply $y^{2}$ by $1$ .

$(2 - y - 2 y + y^{2}) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.2.2

Subtract $2 y$ from $- y$ .

$(2 - 3 y + y^{2}) (- 3 - y) - 24 + 12 y = 0$

Step 1.1.3

Expand $(2 - 3 y + y^{2}) (- 3 - y)$ by multiplying each term in the first expression by each term in the second expression.

$2 \cdot - 3 + 2 (- y) - 3 y \cdot - 3 - 3 y (- y) + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4

Simplify each term.

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Step 1.1.4.1

Multiply $2$ by $- 3$ .

$- 6 + 2 (- y) - 3 y \cdot - 3 - 3 y (- y) + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.2

Multiply $- 1$ by $2$ .

$- 6 - 2 y - 3 y \cdot - 3 - 3 y (- y) + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.3

Multiply $- 3$ by $- 3$ .

$- 6 - 2 y + 9 y - 3 y (- y) + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.4

Rewrite using the commutative property of multiplication.

$- 6 - 2 y + 9 y - 3 \cdot - 1 y \cdot y + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.5

Multiply $y$ by $y$ by adding the exponents.

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Step 1.1.4.5.1

Move $y$ .

$- 6 - 2 y + 9 y - 3 \cdot - 1 (y \cdot y) + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.5.2

Multiply $y$ by $y$ .

$- 6 - 2 y + 9 y - 3 \cdot - 1 y^{2} + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.6

Multiply $- 3$ by $- 1$ .

$- 6 - 2 y + 9 y + 3 y^{2} + y^{2} \cdot - 3 + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.7

Move $- 3$ to the left of $y^{2}$ .

$- 6 - 2 y + 9 y + 3 y^{2} - 3 \cdot y^{2} + y^{2} (- y) - 24 + 12 y = 0$

Step 1.1.4.8

Rewrite using the commutative property of multiplication.

$- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - y^{2} y - 24 + 12 y = 0$

Step 1.1.4.9

Multiply $y^{2}$ by $y$ by adding the exponents.

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Step 1.1.4.9.1

Move $y$ .

$- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - (y \cdot y^{2}) - 24 + 12 y = 0$

Step 1.1.4.9.2

Multiply $y$ by $y^{2}$ .

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Step 1.1.4.9.2.1

Raise $y$ to the power of $1$ .

$- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - (y^{1} y^{2}) - 24 + 12 y = 0$

Step 1.1.4.9.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - y^{1 + 2} - 24 + 12 y = 0$

Step 1.1.4.9.3

Add $1$ and $2$ .

$- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - y^{3} - 24 + 12 y = 0$

Step 1.1.5

Combine the opposite terms in $- 6 - 2 y + 9 y + 3 y^{2} - 3 y^{2} - y^{3}$ .

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Step 1.1.5.1

Subtract $3 y^{2}$ from $3 y^{2}$ .

$- 6 - 2 y + 9 y + 0 - y^{3} - 24 + 12 y = 0$

Step 1.1.5.2

Add $- 6 - 2 y + 9 y$ and $0$ .

$- 6 - 2 y + 9 y - y^{3} - 24 + 12 y = 0$

Step 1.1.6

Add $- 2 y$ and $9 y$ .

$- 6 + 7 y - y^{3} - 24 + 12 y = 0$

Step 1.2

Simplify by adding terms.

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Step 1.2.1

Subtract $24$ from $- 6$ .

$7 y - y^{3} - 30 + 12 y = 0$

Step 1.2.2

Add $7 y$ and $12 y$ .

$19 y - y^{3} - 30 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $- 1$ out of $19 y - y^{3} - 30$ .

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Step 2.1.1

Reorder $19 y$ and $- y^{3}$ .

$- y^{3} + 19 y - 30 = 0$

Step 2.1.2

Factor $- 1$ out of $- y^{3}$ .

$- (y^{3}) + 19 y - 30 = 0$

Step 2.1.3

Factor $- 1$ out of $19 y$ .

$- (y^{3}) - (- 19 y) - 30 = 0$

Step 2.1.4

Rewrite $- 30$ as $- 1 (30)$ .

$- (y^{3}) - (- 19 y) - 1 \cdot 30 = 0$

Step 2.1.5

Factor $- 1$ out of $- (y^{3}) - (- 19 y)$ .

$- (y^{3} - 19 y) - 1 \cdot 30 = 0$

Step 2.1.6

Factor $- 1$ out of $- (y^{3} - 19 y) - 1 (30)$ .

$- (y^{3} - 19 y + 30) = 0$

Step 2.2

Factor $y^{3} - 19 y + 30$ using the rational roots test.

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Step 2.2.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 30, \pm 2, \pm 15, \pm 3, \pm 10, \pm 5, \pm 6$

$q = \pm 1$

Step 2.2.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 30, \pm 2, \pm 15, \pm 3, \pm 10, \pm 5, \pm 6$

Step 2.2.3

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

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Step 2.2.3.1

Substitute $2$ into the polynomial.

$2^{3} - 19 \cdot 2 + 30$

Step 2.2.3.2

Raise $2$ to the power of $3$ .

$8 - 19 \cdot 2 + 30$

Step 2.2.3.3

Multiply $- 19$ by $2$ .

$8 - 38 + 30$

Step 2.2.3.4

Subtract $38$ from $8$ .

$- 30 + 30$

Step 2.2.3.5

Add $- 30$ and $30$ .

$0$

Step 2.2.4

Since $2$ is a known root, divide the polynomial by $y - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{y^{3} - 19 y + 30}{y - 2}$

Step 2.2.5

Divide $y^{3} - 19 y + 30$ by $y - 2$ .

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Step 2.2.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$2$

$y^{3}$

$0 y^{2}$

$19 y$

$30$

Step 2.2.5.2

Divide the highest order term in the dividend $y^{3}$ by the highest order term in divisor $y$ .

					$y^{2}$
	$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$

Step 2.2.5.3

Multiply the new quotient term by the divisor.

				$y^{2}$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			+	$y^{3}$	-	$2 y^{2}$

Step 2.2.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{3} - 2 y^{2}$

				$y^{2}$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$

Step 2.2.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$y^{2}$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$

Step 2.2.5.6

Pull the next terms from the original dividend down into the current dividend.

				$y^{2}$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$

Step 2.2.5.7

Divide the highest order term in the dividend $2 y^{2}$ by the highest order term in divisor $y$ .

				$y^{2}$	+	$2 y$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$

Step 2.2.5.8

Multiply the new quotient term by the divisor.

				$y^{2}$	+	$2 y$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					+	$2 y^{2}$	-	$4 y$

Step 2.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 y^{2} - 4 y$

				$y^{2}$	+	$2 y$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$

Step 2.2.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$y^{2}$	+	$2 y$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$

Step 2.2.5.11

Pull the next terms from the original dividend down into the current dividend.

				$y^{2}$	+	$2 y$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$	+	$30$

Step 2.2.5.12

Divide the highest order term in the dividend $- 15 y$ by the highest order term in divisor $y$ .

				$y^{2}$	+	$2 y$	-	$15$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$	+	$30$

Step 2.2.5.13

Multiply the new quotient term by the divisor.

				$y^{2}$	+	$2 y$	-	$15$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$	+	$30$
							-	$15 y$	+	$30$

Step 2.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 y + 30$

				$y^{2}$	+	$2 y$	-	$15$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$	+	$30$
							+	$15 y$	-	$30$

Step 2.2.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$y^{2}$	+	$2 y$	-	$15$
$y$	-	$2$		$y^{3}$	+	$0 y^{2}$	-	$19 y$	+	$30$
			-	$y^{3}$	+	$2 y^{2}$
					+	$2 y^{2}$	-	$19 y$
					-	$2 y^{2}$	+	$4 y$
							-	$15 y$	+	$30$
							+	$15 y$	-	$30$
										$0$

Step 2.2.5.16

Since the remander is $0$ , the final answer is the quotient.

$y^{2} + 2 y - 15$

Step 2.2.6

Write $y^{3} - 19 y + 30$ as a set of factors.

$- ((y - 2) (y^{2} + 2 y - 15)) = 0$

Step 2.3

Factor.

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Step 2.3.1

Factor $y^{2} + 2 y - 15$ using the AC method.

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Step 2.3.1.1

Factor $y^{2} + 2 y - 15$ using the AC method.

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Step 2.3.1.1.1

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 2.3.1.1.2

Write the factored form using these integers.

$- ((y - 2) ((y - 3) (y + 5))) = 0$

Step 2.3.1.2

Remove unnecessary parentheses.

$- ((y - 2) (y - 3) (y + 5)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$- (y - 2) (y - 3) (y + 5) = 0$

Step 3

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

$y - 2 = 0$

$y - 3 = 0$

$y + 5 = 0$

Step 4

Set $y - 2$ equal to $0$ and solve for $y$ .

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Step 4.1

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Step 4.2

Add $2$ to both sides of the equation.

$y = 2$

Step 5

Set $y - 3$ equal to $0$ and solve for $y$ .

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Step 5.1

Set $y - 3$ equal to $0$ .

$y - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$y = 3$

Step 6

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

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Step 6.1

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

$y + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$y = - 5$

Step 7

The final solution is all the values that make $- (y - 2) (y - 3) (y + 5) = 0$ true.

$y = 2, 3, - 5$