Question

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

Answer 1

`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)`

Answer 2

`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.

Answer 3

`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.