Simplify (p^2-14p+40)/(16p^3)+(3p^2-18p+24)/(2p^2-4p)

$\frac{p^{2} - 14 p + 40}{16 p^{3}} + \frac{3 p^{2} - 18 p + 24}{2 p^{2} - 4 p}$

Simplify each term.

Factor $p^{2} - 14 p + 40$ using the AC method.

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 $40$ and whose sum is $- 14$ .

$- 10, - 4$

Write the factored form using these integers.

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 p^{2} - 18 p + 24}{2 p^{2} - 4 p}$

Simplify the numerator.

Factor $3$ out of $3 p^{2} - 18 p + 24$ .

Factor $3$ out of $3 p^{2}$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p^{2}) - 18 p + 24}{2 p^{2} - 4 p}$

Factor $3$ out of $- 18 p$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p^{2}) + 3 (- 6 p) + 24}{2 p^{2} - 4 p}$

Factor $3$ out of $24$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 p^{2} + 3 (- 6 p) + 3 \cdot 8}{2 p^{2} - 4 p}$

Factor $3$ out of $3 p^{2} + 3 (- 6 p)$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p^{2} - 6 p) + 3 \cdot 8}{2 p^{2} - 4 p}$

Factor $3$ out of $3 (p^{2} - 6 p) + 3 \cdot 8$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p^{2} - 6 p + 8)}{2 p^{2} - 4 p}$

Factor $p^{2} - 6 p + 8$ using the AC method.

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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 ((p - 4) (p - 2))}{2 p^{2} - 4 p}$

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (p - 2)}{2 p^{2} - 4 p}$

Factor $2 p$ out of $2 p^{2} - 4 p$ .

Factor $2 p$ out of $2 p^{2}$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (p - 2)}{2 p (p) - 4 p}$

Factor $2 p$ out of $- 4 p$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (p - 2)}{2 p (p) + 2 p (- 2)}$

Factor $2 p$ out of $2 p (p) + 2 p (- 2)$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (p - 2)}{2 p (p - 2)}$

Cancel the common factor of $p - 2$ .

Cancel the common factor.

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (p - 2)}{2 p (p - 2)}$

Rewrite the expression.

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4)}{2 p}$

To write $\frac{3 (p - 4)}{2 p}$ as a fraction with a common denominator, multiply by $\frac{8 p^{2}}{8 p^{2}}$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4)}{2 p} \cdot \frac{8 p^{2}}{8 p^{2}}$

Write each expression with a common denominator of $16 p^{3}$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 (p - 4)}{2 p}$ by $\frac{8 p^{2}}{8 p^{2}}$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{2 p (8 p^{2})}$

Multiply $8$ by $2$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{16 p \cdot p^{2}}$

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

Move $p^{2}$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{16 (p^{2} p)}$

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

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

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{16 (p^{2} p^{1})}$

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

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{16 p^{2 + 1}}$

Add $2$ and $1$ .

$\frac{(p - 10) (p - 4)}{16 p^{3}} + \frac{3 (p - 4) (8 p^{2})}{16 p^{3}}$

Combine the numerators over the common denominator.

$\frac{(p - 10) (p - 4) + 3 (p - 4) (8 p^{2})}{16 p^{3}}$

Simplify the numerator.

Factor $p - 4$ out of $(p - 10) (p - 4) + 3 (p - 4) (8 p^{2})$ .

Factor $p - 4$ out of $(p - 10) (p - 4)$ .

$\frac{(p - 4) (p - 10) + 3 (p - 4) (8 p^{2})}{16 p^{3}}$

Factor $p - 4$ out of $3 (p - 4) (8 p^{2})$ .

$\frac{(p - 4) (p - 10) + (p - 4) (3 (8 p^{2}))}{16 p^{3}}$

Factor $p - 4$ out of $(p - 4) (p - 10) + (p - 4) (3 (8 p^{2}))$ .

$\frac{(p - 4) (p - 10 + 3 (8 p^{2}))}{16 p^{3}}$

Factor by grouping.

Reorder terms.

$\frac{(p - 4) (3 \cdot 8 p^{2} + p - 10)}{16 p^{3}}$

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 $a \cdot c = 24 \cdot - 10 = - 240$ and whose sum is $b = 1$ .

Multiply $3$ by $8$ .

$\frac{(p - 4) (24 p^{2} + p - 10)}{16 p^{3}}$

Multiply by $1$ .

$\frac{(p - 4) (24 p^{2} + 1 p - 10)}{16 p^{3}}$

Rewrite $1$ as $- 15$ plus $16$

$\frac{(p - 4) (24 p^{2} + (- 15 + 16) p - 10)}{16 p^{3}}$

Apply the distributive property.

$\frac{(p - 4) (24 p^{2} - 15 p + 16 p - 10)}{16 p^{3}}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(p - 4) ((24 p^{2} - 15 p) + 16 p - 10)}{16 p^{3}}$

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

$\frac{(p - 4) (3 p (8 p - 5) + 2 (8 p - 5))}{16 p^{3}}$

Factor the polynomial by factoring out the greatest common factor, $8 p - 5$ .

$\frac{(p - 4) ((8 p - 5) (3 p + 2))}{16 p^{3}}$

$\frac{(p - 4) (8 p - 5) (3 p + 2)}{16 p^{3}}$

Simplify (p^2-14p+40)/(16p^3)+(3p^2-18p+24)/(2p^2-4p)

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