Algebra Examples

$\frac{49 - y^{2}}{6 y^{2} + 42 y} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $49$ as $7^{2}$ .

$\frac{7^{2} - y^{2}}{6 y^{2} + 42 y} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = y$ .

$\frac{(7 + y) (7 - y)}{6 y^{2} + 42 y} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 2

Factor $6 y$ out of $6 y^{2} + 42 y$ .

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Step 2.1

Factor $6 y$ out of $6 y^{2}$ .

$\frac{(7 + y) (7 - y)}{6 y (y) + 42 y} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 2.2

Factor $6 y$ out of $42 y$ .

$\frac{(7 + y) (7 - y)}{6 y (y) + 6 y (7)} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 2.3

Factor $6 y$ out of $6 y (y) + 6 y (7)$ .

$\frac{(7 + y) (7 - y)}{6 y (y + 7)} \cdot \frac{48 y^{3}}{y^{2} - 10 y + 21}$

Step 3

Factor $y^{2} - 10 y + 21$ using the AC method.

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Step 3.1

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 $21$ and whose sum is $- 10$ .

$- 7, - 3$

Step 3.2

Write the factored form using these integers.

$\frac{(7 + y) (7 - y)}{6 y (y + 7)} \cdot \frac{48 y^{3}}{(y - 7) (y - 3)}$

Step 4

Simplify terms.

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Step 4.1

Combine.

$\frac{(7 + y) (7 - y) (48 y^{3})}{6 y (y + 7) ((y - 7) (y - 3))}$

Step 4.2

Cancel the common factor of $48$ and $6$ .

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Step 4.2.1

Factor $6$ out of $(7 + y) (7 - y) (48 y^{3})$ .

$\frac{6 ((7 + y) (7 - y) (8 y^{3}))}{6 y (y + 7) ((y - 7) (y - 3))}$

Step 4.2.2

Cancel the common factors.

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Step 4.2.2.1

Factor $6$ out of $6 y (y + 7) ((y - 7) (y - 3))$ .

$\frac{6 ((7 + y) (7 - y) (8 y^{3}))}{6 ((y (y + 7)) ((y - 7) (y - 3)))}$

Step 4.2.2.2

Cancel the common factor.

$\frac{6 ((7 + y) (7 - y) (8 y^{3}))}{6 ((y (y + 7)) ((y - 7) (y - 3)))}$

Step 4.2.2.3

Rewrite the expression.

$\frac{(7 + y) (7 - y) (8 y^{3})}{(y (y + 7)) ((y - 7) (y - 3))}$

Step 4.3

Cancel the common factor of $y^{3}$ and $y$ .

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Step 4.3.1

Factor $y$ out of $(7 + y) (7 - y) (8 y^{3})$ .

$\frac{y ((7 + y) (7 - y) (8 y^{2}))}{(y (y + 7)) ((y - 7) (y - 3))}$

Step 4.3.2

Cancel the common factors.

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Step 4.3.2.1

Factor $y$ out of $(y (y + 7)) ((y - 7) (y - 3))$ .

$\frac{y ((7 + y) (7 - y) (8 y^{2}))}{y ((y + 7) ((y - 7) (y - 3)))}$

Step 4.3.2.2

Cancel the common factor.

$\frac{y ((7 + y) (7 - y) (8 y^{2}))}{y ((y + 7) ((y - 7) (y - 3)))}$

Step 4.3.2.3

Rewrite the expression.

$\frac{(7 + y) (7 - y) (8 y^{2})}{(y + 7) ((y - 7) (y - 3))}$

Step 4.4

Cancel the common factor of $7 + y$ and $y + 7$ .

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Step 4.4.1

Reorder terms.

$\frac{(y + 7) (7 - y) (8 y^{2})}{(y + 7) ((y - 7) (y - 3))}$

Step 4.4.2

Cancel the common factor.

$\frac{(y + 7) (7 - y) (8 y^{2})}{(y + 7) ((y - 7) (y - 3))}$

Step 4.4.3

Rewrite the expression.

$\frac{(7 - y) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5

Cancel the common factor of $7 - y$ and $y - 7$ .

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Step 4.5.1

Rewrite $7$ as $- 1 (- 7)$ .

$\frac{(- 1 (- 7) - y) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5.2

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

$\frac{(- 1 (- 7) - (y)) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5.3

Factor $- 1$ out of $- 1 (- 7) - (y)$ .

$\frac{- 1 (- 7 + y) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5.4

Reorder terms.

$\frac{- 1 (y - 7) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5.5

Cancel the common factor.

$\frac{- 1 (y - 7) (8 y^{2})}{(y - 7) (y - 3)}$

Step 4.5.6

Rewrite the expression.

$\frac{- 1 (8 y^{2})}{y - 3}$

Step 4.6

Simplify the expression.

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Step 4.6.1

Multiply $- 1$ by $8$ .

$\frac{- 8 y^{2}}{y - 3}$

Step 4.6.2

Move the negative in front of the fraction.

$- \frac{8 y^{2}}{y - 3}$