Question

Simplify the following: `" "a: color(white)("d")(x^2+7x+12)/(x+3)` `" "b: color(white)("d") sqrt((x^9y^4)/x^5)` `" "c: (3^(-1)a^4b^(-3))^2/(6a^2b^(-1)c^(-2))^2` ?

Answer 1

Solution to a) `->x+4`

Explanation:

Solution to a)

Factorising takes quite a bit of practice as you need to build up a 'memory base' that you can draw on.

Given:`" "(x^2+7x+12)/(x+3)`

Consider the top bit (numerator)

Notice that `3xx4=12" and that "3+4=7`

The coefficient of `x^2` (number in front of it) is 1
so we can write `x^2+7x+12" as "(x+3)(x+4)`

Putting it all together we have:

`(x^2+7x+12)/(x+3)" "=" "((x+3)(x+4))/((x+3))`

`" "=" "(x+3)/(x+3)xx(x+4)`

`" "=" "1" "xx(x+4)`

`" "=" "x+4`

Answer 2

Solution to b) `->x^2y^2`

Explanation:

Given:`" "sqrt((x^9y^4)/x^5`

You are looking for squared values as these can be 'taken outside' the root. Also note that (by example) `sqrt(a/b) -> sqrt(a)/sqrt(b)`

Write as: `(sqrt(cancel(x^2)xx cancel(x^2)xx x^2xx x^2xx x xxy^2xxy^2))/(sqrt(cancel(x^2)xx cancel(x^2)xx x)`

`=(x^2y^2cancel(sqrt(x)))/(cancel(sqrt(x)))`

`=x^2y^2`

Answer 3

Solution to c) `" "(b^8c^4)/(4a^16)`

Explanation:

By example:
Note that `a^(-1)=1/a"; "a^(-2)=1/a^2"; "a^(-2/3)=1/(root(3)(a^2))`

Note that `1/(a^(-1))=a"; "1/a^(-2)=a^2"; "1/a^(-2/3)=root(3)(a^2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Consider the numerator:")`
`" "(3^(-1)a^4b^(-3))^(-2)" "->" " (1/3xxa^4xx1/b^3)^(-2)`

`" " = " "(a^4/(3b^3))^(-2)`

`" " = " "((3b^3)/a^4)^(2)`

`" " = " "(9b^6)/a^(12)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider the denominator:")`
`" "(6a^2b^(-1)c^(-2))^2->(6xxa^2xx1/bxx1/c^2)^2`

`" "=" "((6a^2)/(bc^2))^2`

`" "=" "(36a^4)/(b^2c^4)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`color(brown)("numerator " -:" denominator " ->" " (9b^6)/(a^12) -: (36a^4)/(b^2c^4))`

`" " (9b^6)/(a^12) xx (b^2c^4)/(36a^4)`

`" "=" "9/36xx(b^8c^4)/a^16`

`" "=" "(b^8c^4)/(4a^16)`