Algebra Examples

$\frac{3 x^{2} - 28 x - 20}{2 x + 16}$

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 20 = - 60$ and whose sum is $b = - 28$ .

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Step 1.1.1

Factor $- 28$ out of $- 28 x$ .

$\frac{3 x^{2} - 28 (x) - 20}{2 x + 16}$

Step 1.1.2

Rewrite $- 28$ as $2$ plus $- 30$

$\frac{3 x^{2} + (2 - 30) x - 20}{2 x + 16}$

Step 1.1.3

Apply the distributive property.

$\frac{3 x^{2} + 2 x - 30 x - 20}{2 x + 16}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$\frac{(3 x^{2} + 2 x) - 30 x - 20}{2 x + 16}$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{x (3 x + 2) - 10 (3 x + 2)}{2 x + 16}$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$\frac{(3 x + 2) (x - 10)}{2 x + 16}$

Step 2

Factor $2$ out of $2 x + 16$ .

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Step 2.1

Factor $2$ out of $2 x$ .

$\frac{(3 x + 2) (x - 10)}{2 (x) + 16}$

Step 2.2

Factor $2$ out of $16$ .

$\frac{(3 x + 2) (x - 10)}{2 x + 2 \cdot 8}$

Step 2.3

Factor $2$ out of $2 x + 2 \cdot 8$ .

$\frac{(3 x + 2) (x - 10)}{2 (x + 8)}$

Step 3

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes

Step 4