# polynomial: simplify and evaluate (22x5y-3z4/40x-2y-8z5)3

#### janekommer

##### New member
I need to Simplify and evaluate this problem and am stuck as to where to begin.

(22x5y-3z4/40x-2y-8z5)3

#### galactus

##### Super Moderator
Staff member
I am going out on a limb here. Do you mean?:

$$\displaystyle \L\\\left(\frac{22x^{5}y-3z^{4}}{40x-2y-8z^{5}}\right)^{3}$$

If this is the case, please use approriate sybols to represent 'powers', such as ^.

In general, few are going to know what you mean and, therefore, you won't get an accurate response.

Most would think you mean:

$$\displaystyle \L\\3(\frac{22x\cdot{5y}-4z\cdot{3}}{40x-2y-8z\cdot{5}})$$

Unless of course, that is what you mean. Which seems unlikely.

#### stapel

##### Super Moderator
Staff member
Note: Until you are given values for the variables, it is impossible to do the "evaluate" part of this exercise. Sorry.

Eliz.

#### janekommer

##### New member
Sorry it should read. (2^2x^5y^-3z^4/4^0x^-2y^-8z^5)^3

#### stapel

##### Super Moderator
Staff member
janekommer said:
Sorry it should read. (2^2x^5y^-3z^4/4^0x^-2y^-8z^5)^3
So you mean the following...?

. . . . .$$\displaystyle \L \left(\frac{2^2\, x^5\, y^{-3}\, z^4}{4^0\, x^{-2}\, y^{-8}\, z^5}\right)^3$$

Then this is a simple matter of taking the cube through, and simplifying, using the addition and subtraction rules you memorized for exponents, namely:

. . . . .(x<sup>m</sup>)(x<sup>n</sup>) = x<sup>m+n</sup>

. . . . .(x<sup>m</sup>) / (x<sup>n</sup>) = x<sup>m-n</sup>

. . . . .x<sup>-m</sup> = 1/x<sup>m</sup>

Where are you stuck? Please reply showing all of your steps so far. For instance, you squared the 2, converted 4<sup>0</sup> to 1, and... then what?

Thank you.

Eliz.

#### Denis

##### Senior Member
Also, (a^m)^n = a^(m*n) : use that when doing the cubing.

#### janekommer

##### New member
Here is where I am stuck. Do I divide or add these up.

If I divide here is where I am.

2^2/4^0=1
X^5/x^-2=x^-2.5
Y^-3/y^-8=y^.375
Z^4/z^5=z^20

adding would look like this

2^2+4^0=6^2
X^5/x^-2=x^3
Y^-3/y^-8=y^-11
Z^4/z^5=z^9

W

#### wantedcriminal03

##### Guest
X^0=1
where x is any number

so 2^2/4^0 = 4/1=4

when dividing x^a/x^b you just subtract the powers so ypu get x^(a-b)

so X^5/x^-2 = x^(5--2) = x^7

and just do that for y and z and multiply it all together and you got the answer.

#### galactus

##### Super Moderator
Staff member
You need to learn the exponent laws and it'll fall into place.

See that 3 on the outside of the 'big' parentheses?. Multiply all your exponents inside by it. You get:

$$\displaystyle \L\\\left(\frac{4x^{5}y^{-3}z^{4}}{x^{-2}y^{-8}z^{5}}\right)^{3}$$

Becomes:

$$\displaystyle \L\\\frac{64x^{15}y^{-9}z^{12}}{x^{-6}y^{-24}z^{15}}$$

Rearrange so they're all positive exponents(if you want):

$$\displaystyle \L\\\frac{64x^{15}z^{12}x^{6}y^{24}}{y^{9}z^{15}}$$

Remember, when you multiply add exponents, and when you divide subtract exponents:

$$\displaystyle \H\\\frac{64x^{21}y^{15}}{z^{3}}$$