Algebra Examples

$\frac{2 x^{2} - 3 x + 1}{2 x + 16}$

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

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Step 1.1.1

Factor $- 3$ out of $- 3 x$ .

$\frac{2 x^{2} - 3 (x) + 1}{2 x + 16}$

Step 1.1.2

Rewrite $- 3$ as $- 1$ plus $- 2$

$\frac{2 x^{2} + (- 1 - 2) x + 1}{2 x + 16}$

Step 1.1.3

Apply the distributive property.

$\frac{2 x^{2} - 1 x - 2 x + 1}{2 x + 16}$

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.

$\frac{(2 x^{2} - 1 x) - 2 x + 1}{2 x + 16}$

Step 1.2.2

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

$\frac{x (2 x - 1) - (2 x - 1)}{2 x + 16}$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{(2 x - 1) (x - 1)}{2 x + 16}$

Step 2

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

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Step 2.1

Factor $2$ out of $2 x$ .

$\frac{(2 x - 1) (x - 1)}{2 (x) + 16}$

Step 2.2

Factor $2$ out of $16$ .

$\frac{(2 x - 1) (x - 1)}{2 x + 2 \cdot 8}$

Step 2.3

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

$\frac{(2 x - 1) (x - 1)}{2 (x + 8)}$

Step 3

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

No holes

Step 4