Question

Factor completely 12t^8 - 75t^4

Answer 1

plase put in evidence 3t^4, first

Answer 2

you should get this:
\[3t ^{4}(4t ^{4}-25)=3t ^{4}(2t ^{2}-5)(2t ^{2}+5)=\]
\[=3t ^{4}(\sqrt{2}t-\sqrt{5})(\sqrt{2}t+\sqrt{5})(2t ^{2}+5)\]

Answer 3

I don't get the last oart

Answer 4

Part *

Answer 5

the last part, is because:
\[2t ^{2}-5\]
is a difference between two square quantities, namely:
\[\sqrt{2}t, \sqrt{5}\]
so you have to apply the subsequent formula, as you well know:
\[a ^{2}-b ^{2}=(a+b)(a-b)\]
where:
\[a=\sqrt{2}t, b=\sqrt{5}\]

Answer 6

I never leArned that this is algebra 2 right?

Answer 7

I think so, even if you can stop at first step

Answer 8

What about this ques.
Express the product of (1/2y^2 -1/3y) and (12y+3/5) as a trinomial

Answer 9

is your product this?
\[(\frac{ 1 }{ 2 }y ^{2}-\frac{ 1 }{ 3 }y)(12y+\frac{ 3 }{ 5 })\]

Answer 10

Yes!

Answer 11

ok! you have to apply the distributive property of multiplication over the sum:

Answer 12

sorry I rewrite better:
\[\frac{ y ^{2} }{ 2 }*\left( 12y+\frac{ 3 }{ 5 } \right)-\frac{ y }{ 3 }*\left( 12y+\frac{ 3 }{ 5 } \right)=\]

Answer 13

so, we have:

\[\frac{ y ^{?} }{ 2 }*12y +\frac{ y ^{2} }{ 2 }*\frac{ 3 }{ 5 }-\frac{ y }{ 3 }*12y-\frac{ y }{ 3 }*\frac{ 3 }{ 5 } = \]

Answer 14

sorry the first term is:
\[\frac{ y ^{2} }{ 2 }*12y\]

Answer 15

s after some simplification we get
\[6y ^{3}-\frac{ 37 }{ 10 }y ^{2}-\frac{ y }{ 5 }\]
@Sarah_s

Answer 16

So that would be the answer?

Answer 17

yes, is it right?

Answer 18

Okay and for this one
If f(x)=x^2 -6, find f^-1(x)
Is the answer y=x^2+6?

Answer 19

please do you want to calculate:
\[\frac{ 1 }{ f(x) }\]

Answer 20

No it's f^-1(x) so the inverse

Answer 21

ok, then inverse function is:
\[x=\pm \sqrt{y+6}\]

Answer 22

How

Answer 23

please insert x from the above formula into expression for f(x) and you will get y

Answer 24

Wouldn't the inverse be x^2-6 then get in y = for! by +6?