Algebra Examples

$p (x) = 2 x^{3} + 4 x^{2} + 6 x + 7$ , $x = \frac{1}{2}$

Step 1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$p (\frac{1}{2}) = 2 {(\frac{1}{2})}^{3} + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2

Simplify the result.

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the product rule to $\frac{1}{2}$ .

$p (\frac{1}{2}) = 2 (\frac{1^{3}}{2^{3}}) + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.2

One to any power is one.

$p (\frac{1}{2}) = 2 (\frac{1}{2^{3}}) + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.3

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

$p (\frac{1}{2}) = 2 (\frac{1}{8}) + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.4

Cancel the common factor of $2$ .

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Step 2.1.4.1

Factor $2$ out of $8$ .

$p (\frac{1}{2}) = 2 (\frac{1}{2 (4)}) + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.4.2

Cancel the common factor.

$p (\frac{1}{2}) = 2 (\frac{1}{2 \cdot 4}) + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.4.3

Rewrite the expression.

$p (\frac{1}{2}) = \frac{1}{4} + 4 {(\frac{1}{2})}^{2} + 6 (\frac{1}{2}) + 7$

Step 2.1.5

Apply the product rule to $\frac{1}{2}$ .

$p (\frac{1}{2}) = \frac{1}{4} + 4 (\frac{1^{2}}{2^{2}}) + 6 (\frac{1}{2}) + 7$

Step 2.1.6

One to any power is one.

$p (\frac{1}{2}) = \frac{1}{4} + 4 (\frac{1}{2^{2}}) + 6 (\frac{1}{2}) + 7$

Step 2.1.7

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

$p (\frac{1}{2}) = \frac{1}{4} + 4 (\frac{1}{4}) + 6 (\frac{1}{2}) + 7$

Step 2.1.8

Cancel the common factor of $4$ .

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Step 2.1.8.1

Cancel the common factor.

$p (\frac{1}{2}) = \frac{1}{4} + 4 (\frac{1}{4}) + 6 (\frac{1}{2}) + 7$

Step 2.1.8.2

Rewrite the expression.

$p (\frac{1}{2}) = \frac{1}{4} + 1 + 6 (\frac{1}{2}) + 7$

Step 2.1.9

Cancel the common factor of $2$ .

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Step 2.1.9.1

Factor $2$ out of $6$ .

$p (\frac{1}{2}) = \frac{1}{4} + 1 + 2 (3) (\frac{1}{2}) + 7$

Step 2.1.9.2

Cancel the common factor.

$p (\frac{1}{2}) = \frac{1}{4} + 1 + 2 \cdot (3 (\frac{1}{2})) + 7$

Step 2.1.9.3

Rewrite the expression.

$p (\frac{1}{2}) = \frac{1}{4} + 1 + 3 + 7$

Step 2.2

Find the common denominator.

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Step 2.2.1

Write $1$ as a fraction with denominator $1$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{1}{1} + 3 + 7$

Step 2.2.2

Multiply $\frac{1}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{1}{1} \cdot \frac{4}{4} + 3 + 7$

Step 2.2.3

Multiply $\frac{1}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + 3 + 7$

Step 2.2.4

Write $3$ as a fraction with denominator $1$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3}{1} + 7$

Step 2.2.5

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3}{1} \cdot \frac{4}{4} + 7$

Step 2.2.6

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3 \cdot 4}{4} + 7$

Step 2.2.7

Write $7$ as a fraction with denominator $1$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3 \cdot 4}{4} + \frac{7}{1}$

Step 2.2.8

Multiply $\frac{7}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3 \cdot 4}{4} + \frac{7}{1} \cdot \frac{4}{4}$

Step 2.2.9

Multiply $\frac{7}{1}$ by $\frac{4}{4}$ .

$p (\frac{1}{2}) = \frac{1}{4} + \frac{4}{4} + \frac{3 \cdot 4}{4} + \frac{7 \cdot 4}{4}$

Step 2.3

Combine the numerators over the common denominator.

$p (\frac{1}{2}) = \frac{1 + 4 + 3 \cdot 4 + 7 \cdot 4}{4}$

Step 2.4

Simplify each term.

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Step 2.4.1

Multiply $3$ by $4$ .

$p (\frac{1}{2}) = \frac{1 + 4 + 12 + 7 \cdot 4}{4}$

Step 2.4.2

Multiply $7$ by $4$ .

$p (\frac{1}{2}) = \frac{1 + 4 + 12 + 28}{4}$

Step 2.5

Simplify by adding numbers.

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Step 2.5.1

Add $1$ and $4$ .

$p (\frac{1}{2}) = \frac{5 + 12 + 28}{4}$

Step 2.5.2

Add $5$ and $12$ .

$p (\frac{1}{2}) = \frac{17 + 28}{4}$

Step 2.5.3

Add $17$ and $28$ .

$p (\frac{1}{2}) = \frac{45}{4}$

Step 2.6

The final answer is $\frac{45}{4}$ .

$\frac{45}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{45}{4}$

Decimal Form:

$11.25$

Mixed Number Form:

$11 \frac{1}{4}$

Step 4