What is (80x^4 + 48x^3 + 80x^2) -: 8x?

1 Answer
Feb 18, 2016

(80x^4 + 48x^3 + 80x^2)/(8x) = 10x^3 + 6x^2 + 10x

Explanation:

Starting with: (80x^4 + 48x^3 + 80x^2)/(8x)

Dividing something by 8x is the same as multiplying it by (8x)^-1. Thus we have:
(80x^4 + 48x^3 + 80x^2)(8x)^-1

Notice that each of the terms is a factor of 8:
((8^1)(10)x^4 + (8^1)(6)x^3 + (8^1)(10)x^2)(8^-1x^-1)
Now multiply out the outer brackets and use the property (x^a)(x^b)=x^(a+b):
(8^(1-1))(10)x^(4-1) + (8^(1-1))(6)x^(3-1) + (8^(1-1)(10)x^(2-1))
(8^0)(10)x^3 + (8^0)(6)x^2 + (8^0)(10)x^1)

Noting that x^0 = 1 and x^1=x:
(10)x^3 + (6)x^2 + (10)x^1
10x^3 + 6x^2 + 10x

If you didn't spot that all terms were a factor of 8, you could also say that 8^-1 = 1/8 = 0.125 and multiply each of the terms in the left bracket by 0.125/x.

Alternatively, you may find it easier to split the terms up:
((80x^4)/(8x) + (48x^3)/(8x) + (80x^2)/(8x)) = 10x^3 + 6x^2 + 10x