Question

Write these as a product

a) (2b–5)^2–36
b) 9–(7+3a)^2
c) (4–11m)^2–1

thanks in advance :)

Answer 1

a)

`{(2b - 5)}^(2) - 36 =( 2b - 11)(2b + 1)`

b)

`9 - {(7 + 3a)}^(2) = (3a - 4)(3a + 11)`

c)

`( {4 - 11m)}^(2) - 1 =( 3 - 11m )( 5 - 11m )`

Step-by-step explanation

a) The given expresion is

`{(2b - 5)}^(2) - 36`

We rewrite as difference of two squares

`{(2b - 5)}^(2) - 36 = {(2b - 5)}^(2) - {6}^(2)`

Recall that:

`{x}^(2) - {y}^(2) = (x + y)(x - y)`

This implies that:

`{(2b - 5)}^(2) - 36 =( {(2b - 5)} -6)(2b - 5 )+ 6)`

Or

`{(2b - 5)}^(2) - 36 =( 2b - 5-6)(2b - 5 + 6)`

This simplifies to give:

`{(2b - 5)}^(2) - 36 =( 2b - 11)(2b + 1)`

b) The second expression is

`9 - {(7 + 3a)}^(2)`

We rewrite as perfect squares yo get:

`9 - {(7 + 3a)}^(2) = {3}^(2) - {(7 + 3a)}^(2)`

This gives:

`9 - {(7 + 3a)}^(2) = ({3} - {(7 + 3a)})({3} + {(7 + 3a)})`

This implies that

`9 - {(7 + 3a)}^(2) = ({3} - 7 + 3a)({3} + 7 + 3a)`

We simplify to get:

`9 - {(7 + 3a)}^(2) = (3a - 4)(3a + 11)`

c) The third expression is:

`( {4 - 11m)}^(2) - 1`

We obtain the difference of two squares as:

`( {4 - 11m)}^(2) - 1 =( ( {4 - 11m)} - 1 )( ( {4 - 11m)} + 1 )`

We simplify within the parenthesis to get:

`( {4 - 11m)}^(2) - 1 =( 4 - 11m - 1 )( 4 - 11m+ 1 )`

We simplify further to get;

`( {4 - 11m)}^(2) - 1 =( 3 - 11m )( 5 - 11m )`