How do you simplify #\frac { 5a ^ { 2} - 45} { 3a ^ { 2} - 12a } \div \frac { a ^ { 3} + 3a ^ { 2} } { 4a ^ { 2} - 16a } \cdot \frac { 21a ^ { 3} + 6a ^ { 2} } { 4a ^ { 2} - 12a }#?

Question

Elijah Allen · Accepted Answer

#(5(7a+2))/a#

#(5a^2-45)/(3a^2-12a) * (4a^2-16a)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

(dividing by a fraction means multiplying by its reciprocal)

#3a^2-12a = 3a(a-4)# #4a^2-16a = 4a(a-4)#

#(5a^2-45)/(3acancel((a-4))) * (4acancel((a-4)))/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#=(5a^2-45)/(3a) * (4a)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#=(5a^2-45)/(3) * (4)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#a^3+3a^2 = a^2(a+3)# #5a^2-45 = 5(a^2-9) = 5(a+3)(a-3)#

#(5cancel((a+3))(a-3))/(3) * (4)/(a^2cancel((a+3)))*(21a^3+6a^2)/(4a^2-12a)#

#=(5(a-3))/(3) * (4)/(a^2)*(21a^3+6a^2)/(4a^2-12a)#

#4a^2-12a = 4a(a-3)#

#(5cancel((a-3)))/(3) * (4)/(a^2)*(21a^3+6a^2)/(4acancel((a-3))#

#=(5)/(3) * (4)/(a^2)*(21a^3+6a^2)/(4a)#

#21a^3+6a^2 = 3a^2(7a+2)#

#(5)/(cancel3) * (4)/(cancel(a^2))*(cancel3cancel(a^2)(7a+2))/(4a)#

#=5/1 * 4/1 * (7a+2)/(4a)#

#5/1 * cancel4/1 * (7a+2)/(cancel4a) = (5*(7a+2))/a#

#=(5(7a+2))/a#

Cora Alexander · Answer

undefined To simplify the given expression:

$$ \frac{{5a^2 - 45}}{{3a^2 - 12a}} \div \frac{{a^3 + 3a^2}}{{4a^2 - 16a}} \cdot \frac{{21a^3 + 6a^2}}{{4a^2 - 12a}} $$

First, we'll factor where possible:

$$ \frac{{5(a^2 - 9)}}{{3a(a - 4)}} \div \frac{{a^2(a + 3)}}{{4a(a - 4)}} \cdot \frac{{3a^2(7a + 2)}}{{4a(a - 3)}} $$

Now, we'll rewrite the division as multiplication by the reciprocal:

$$ \frac{{5(a^2 - 9)}}{{3a(a - 4)}} \cdot \frac{{4a(a - 4)}}{{a^2(a + 3)}} \cdot \frac{{4a(a - 3)}}{{3a^2(7a + 2)}} $$

Now, we'll cancel out common factors:

$$ \frac{{5\cancel{(a^2 - 9)} \cdot 4\cancel{a}\cancel{(a - 4)} \cdot 4\cancel{a}(a - 3)}}{{3\cancel{a}\cancel{(a - 4)} \cdot \cancel{4a}\cancel{(a + 3)} \cdot 3\cancel{a^2}\cancel{(7a + 2)}}} $$

$$ = \frac{{5 \cdot 4 \cdot 4(a - 3)}}{{3 \cdot 3}} $$

$$ = \frac{{320(a - 3)}}{9} $$

So, the simplified expression is $ \frac{{320(a - 3)}}{9} $.