How do you multiply #(-b^2 + 2) /(-b^4 - 6b^2 + 16) * 20/(35b + 70)#?

Question

1100 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T08:05:37

#=4/(7(b^2+8)(b+2))#

#color(blue)(((-b^2 + 2)) /((-b^4) - 6b^2 + 16)) * 20/(35b + 70)#?

Divide #-1# out of the brackets to change the signs

#=color(blue)((-(b^2 - 2)) /(-(b^4 + 6b^2 - 16))) * 20/(35b + 70)#

Negative divided by a negatives gives a positive

Factorise

#=+((b^2 - 2)) /((b^4 + 6b^2 - 16)) * 20/(35(b + 2)#

#=((b^2 - 2)) /((b^2 +8)(b^2-2)) * 20/(35(b+2))" "larr#

#=(cancel((b^2 - 2))) /((b^2 +8)cancel((b^2-2))) * cancel20^4/(cancel35^7(b+2))" "larr# cancel

#=4/(7(b^2+8)(b+2))#