Question

Multiply: z squared plus 6 times z plus 5 all over 3. • 9 times x all over the quantity z squared plus 5 times z, end quantity.

Answer 1

Can someone please help? There's a few more like it I don't understand..

Answer 2

\[\frac{z^2+6z+5}{3}*\frac{9x}{z^2+5z}\]

Answer 3

do you know how to fator Pedro?

Answer 4

factor*

Answer 5

Somewhat. I use online school and it doesn't teach very well, but I'll give it a shot.

Answer 6

try factoring the top of the first and the bottom of the second

Answer 7

would it be (5 + z)(1 + z)?

Answer 8

For the top?

Answer 9

yes and the botom of the second has just a GCF

Answer 10

So it would stay 3?

Answer 11

just leave the bottom as 3. you just want to expand the problem first

Answer 12

Ok, makes sense so far :) Would you factor the second half next?

Answer 13

so for
\[z^2+5z\]
what is your Greatest common factor, or what can you pull out

Answer 14

the 5?

Answer 15

not quite, what does
\[z^2\] and \[5z\] have in common

Answer 16

the z

Answer 17

so you would combing the two?

Answer 18

yes so the gcf is z, if you pull z out of each what are you left with

z(? + ?)

Answer 19

um, 3 and 2?

Answer 20

so greatest common factors work sorta like factoring only you don't have a 3terms so like

\[x^2+15x\]

you can see there is an x in both of them so if you pul out an x, what are you left with

well
\[x*x=x^2\]
and\[15*x=15x\]
so if you get rid of one x in both of these and bring it out front you end up with

\[x(x+15)\]
if you were to multiply it back out you'd get what you originally had

Answer 21

so for yours

\[z*z=z^2\]
\[5*z=5z\]
if you do the same that i did above what do you get

Answer 22

z(x^2+5z) ?

Answer 23

z^2 sorry

Answer 24

if you multiplied that back out would you get the same thing you had before?

Answer 25

you forgot to get rid of the z you pulled out of the original equation =o

Answer 26

so basically to check if you did it correctly use the distibutive property

\[a(b+c)=ab+ac\]
\[z(z^2+5z)=z^3+5z^2\] o no! thats not what you hadbefore!

Answer 27

\[z*z=z^2\]
\[5*z=5z\] if you pull a z out of each

\[z(z+5)=z^2+5\]

Answer 28

5z*

Answer 29

so now you have

\[\frac{(z+1)(z+5)}{3}*\frac{9x}{z(z+5)}\]

Answer 30

when multiplying what can you do with like terms in the numerator and denominator?

Answer 31

Oh, so then you would cancel the like terms out!

Answer 32

yes

Answer 33

z+5 can cancel out, right?

Answer 34

\[\frac{z+1}{3}*\frac{9x}{z}\] then multiply them

Answer 35

yes

Answer 36

you'll get

\[\frac{(z+1)9x}{3z}\]

Answer 37

would there be restrictions?

Answer 38

like z can't equal 0 or -5?

Answer 39

umm the denominator cannot equal zero so the only time that it would equal zero is if z itself was zero

Answer 40

ok and there isn't more simplifying? Like the 3z and z+1?