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# ?

3 Answers
Apr 4, 2017

Answer:

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#

Apr 4, 2017

Answer:

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#

Apr 4, 2017

Answer:

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)#