The number following the ^ is an exponent.
(y^9)(2y^2)^3
2 Answers
Recall three of the laws of exponents I assume your teacher must have covered in your class:
i) xmxn = xm+n
ii) (xm)n = xmn
iii) (xy)n = xnyn
These two laws can be used here:
y9(2y2)3 = 23y9y2*3 = 8y9y6 = 8y9+6 = 8y15
Audy,
(y^9) the first term is y to the 9th power
(2y^2) is (y squared ) times 2. Note that the exponent 2 does not apply to the factor 2.
(2y^2)^3 is the previous term raised to the 3rd power. To see exactly how this calculates expand it:
= 4*y^4 * (2y^2)
= 8*y^6
Go back and include the first term:
(y^9)(2y^2)^3 = (y^9) * 8*y^6
(y^9)(2y^2)^3 = 8*y^15
There are two keys to this problem:
Exponents add when their base numbers are multiplied.
Observe which factors are raised to powers and which are not.
BruceS


