How do you factor #5x^3+6x^2-45x-54#?
1 Answer
May 22, 2016
(5x +6)(x-3)(x+3)
Explanation:
Group the terms in 'pairs'
thus
#[5x^3+6x^2]+[-45x-54]# now factorise each pair
#rArrx^2(5x+6)-9(5x+6)# We now have a common factor of (5x + 6) and taking it out leaves.
#(5x+6)(x^2-9)# Now
#x^2-9" is a difference of squares"# In general a difference of squares factorises as.
#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))# here
#x^2=(x)^2" and " 9=(3)^2rArra=x ,b=3#
#rArrx^2-9=(x-3)(x+3)# Putting this altogether to obtain.
#5x^3+6x^2-45x-54=(5x+6)(x-3)(x+3)#
