How do you multiply #(6p+7)/(p+2)div(36p^2-49)#?

3 Answers

Answer:

# 1/(6p^2 -5p-14 # This is a division problem not a multiplication problem.

Explanation:

# ( 6p +7) / (p+2)# is a division problem written as a fraction.

Then divided by a "whole number " fraction # 36p^2 - 49)/1#

Set the problem up as complex fraction

# (6p +7) / (p+2)/ (36p^2 - 49) / 1 ) #

To make the second fraction (bottom fraction or denominator) disappear multiply both fractions by the multiplicative inverse.

# (6p + 7) / (p+2) xx 1/ ( 36p^2 -49)# and

#( 36p^2 - 49)/1 xx 1/(36p^2 -49) # It is necessary to multiply both the nominator and the denominator by the same "number" or quantity. This is the equality property of multiplication.

# (36p^2 -49)/1 xx 1 / (36p^2 -49)# = 1

The denominator disappears leaving

# ( 6p +7)/ (p+2) xx 1/ (36p^2 - 49)#

# (36p^2 - 49)# is the difference between two squares so

# (36p^2 - 49)# = # ( 6p + 7) xx ( 6p -7)#

Now you have

# ( 6p +7)/{ (p+2) xx (6p +7) xx 6p -7)} #

6p +7 on the top divided by the 6p + 7 on the bottom equals 1

leaving

# 1 / { (p +2) xx ( 6p - 7)}#

Now multiply # (p + 2) xx ( 6p -7 ) #

This gives # 6p^2 + 5p -14# so the answer is

# 1/ (6p^2 +5p - 14) #

Sep 1, 2016

Answer:

#1/((p+2)(6p-7))#

Explanation:

The key is to recognize that #36 p^2 - 49 = (6p+7)(6p-7)# and that
the #6p+7# will cancel the one on the numerator. That leaves #(p+2)(6p-7)# in the denominator.

Sep 1, 2016

Answer:

#1/((p+2)(6p-7))#

Explanation:

#(6p+7)/(p+2)divcolor(red)((36p^2-49))/1#

To divide by a fraction, change #div to xx# and invert the fraction.

=#(6p+7)/(p+2) xx1/color(red)((36p^2-49)) larr " factorise " color(red)("diff of two squares")#

=#cancel((6p+7))/((p+2)) xx1/color(red)(cancel((6p+7))(6p-7)) larr# cancel

=#1/((p+2)(6p-7)) larr# can be left in this form.