How do you multiply #(2x^2+8)div(x^4-16)/(x^2+x-6)#?

1 Answer
Dec 16, 2017

Answer:

#(x+2)/(2(x+3))#

Explanation:

Take into account the formula:
#a^2-b^2=(a+b)(a-b)#

#x^4-16=x^4-4^2=(x^2-4)(x^2+4)#

#(2x^2+8)÷(x^4−16)/(x^2+x−6)=2(x^2+4)÷((x^2-4)(x^2+4))/(x^2+x−6)#

#1/(2cancel((x^2+4)))*((x^2-4)cancel((x^2+4)))/(x^2+x−6)#

We can apply this formula one more for #(x^2-4)#
We can also do something with #(x^2+x−6)#
formula for quadratic function: #ax^2+bx+cquad "or" quad(x-x_1)(x-x_2)#
where #x_1# and #x_2# are roots if it's an equation
We can use the following method only when a=1. In this particular example b=1 and c=-6
in order to get solution we have to: #b=x_1+x_2quad and quadc=x_1*x_2#
The only way how to get 6 is by multiplying 61 or 32 then we have to resolve signs from the second equation

There is no way we can arrange 6 and 1 to get 1. so the solution is 3 and 2 and it's:
#(x^2+x−6)=(x+3)(x-2)#

#1=3+(-2)quad and quad-6=3*(-2)# it's correct

Now we can rewrite:

#1/(2)*((x+2)(x-2))/((x+3)(x−2))=((x+2)cancel((x-2)))/(2(x+3)cancel((x−2)))=(x+2)/(2(x+3))#