Question

Can someone double check my work?

Multiply and Simplify Expression:

(3s - 3)(4s - 5)

Answer 1

\[12s^2 - 15s - 12s + 15\]

= \[12s^2 - 27s + 15\]

Answer 2

\[\huge \checkmark\]

Answer 3

thanks

Answer 4

oh wait, sorry
\[\huge \checkmark\checkmark\]

Answer 5

^.^

Answer 6

So you double checked it... hehe. But you do that if it is right or wrong! =P

Though, she was right.

Answer 7

Go figure :)

Answer 8

That woul be the lemniscate? The go figure?

Answer 9

Lemniscate?
\[\huge \color{blue}\infty\]

Answer 10

For this next one it says find the product, but i did the same steps:

(4a + 5)(4a - 5)

\[16a^2 - 20a + 20a - 25\]

\[16a^2 - 25\]

Answer 11

Yup, @angelina22309 \[\huge \checkmark\]
You'll find that
\[\huge (a+b)(a-b)=a^2-b^2\]

Answer 12

Yah, you go on the lemniscate and tell me when you find the end. Hehe.

And Great angelina22309!

Answer 13

oh the cleverness of me :)

Answer 14

this one is a bit confusing, can you guys help?

Multiply
\[(x^4 + y^3)(x^4 - y^3) = x^8 -x^4y^3 + y^3x^4 - y^6\]

Answer 15

Take a closer look at those middle two terms. Remember that xy=yx.

Answer 16

Also, - means -1, so -xy=-1xy=-1yx

Answer 17

so the middle terms

\[-x^4x^3 +x^3y^4 = -x^7y^7 ?\]

@e.mccormick

Answer 18

Not quite. There is no multiplication there, so this is a negative added to a positive. Another term for tha is subtraction.

Answer 19

Can you show me? I dont understand @e.mccormick

Answer 20

Sure. You put this last:\[(x^4 + y^3)(x^4 - y^3) = x^8 -x^4y^3 + y^3x^4 - y^6\]I am saying change it to:\[(x^4 + y^3)(x^4 - y^3) = x^8 -y^3x^4 + y^3x^4 - y^6\]or\[(x^4 + y^3)(x^4 - y^3) = x^8 + y^3x^4-y^3x^4 - y^6\]
Now, you have those center two terms: \(y^3x^4 -y^3x^4 \). What does that become?

Answer 21

it would be:

\[x^8 - y^6\]

Answer 22

@e.mccormick

Answer 23

You keep addressing them as if they are not a single multiplied term. Let me show you with what PeterPan pointed out:
\[(a+b)(a-b)=a^2-b^2\]
Well, if \(a=x^4\) and \(b=y^3\) then you get:
\[(x^4+b^2)(x^4-b^2)=(x^4)^2-(y^3)^2=x^8-y^6\]
What you have been missng is that those two center terms are the same, but their sign is different. Therefore they cancel.

Answer 24

Once I wrote the answer I noticed that. Im not sure why I did not catch it when I started. @e.mccormick

Answer 25

Yah, sometimes we just get stuck on an idea or forget some key fact and then we can't get it right. I worked for hours on smething trying to make it match the book answer because I thought mine was different. I never realized I had just brought things to the different side of the = sign, so my answer was both sides with opposite signs, which is the same value as what was in the book! And I could not see it!

Answer 26

I'm going to post a fraction one (i always struggle with fractions) in another post. If you could help that would be great

Answer 27

kk.