Hello,
To simplify the expression (54u^2v + 18uv^2)(9u + 3v), we can use the distributive property of multiplication.
First, let's distribute 9u to each term inside the first parentheses: TellPopeyes
9u * 54u^2v = 486u^3v
9u * 18uv^2 = 162u^2v^2
Next, let's distribute 3v to each term inside the first parentheses:
3v * 54u^2v = 162uv^3
3v * 18uv^2 = 54uv^3
Now, we can combine like terms:
486u^3v + 162u^2v^2 + 162uv^3 + 54uv^3
To simplify this further, we can group the terms with the same variables:
(486u^3v + 162u^2v^2) + (162uv^3 + 54uv^3)
Inside each group, we can factor out common terms:
486u^3v + 162u^2v^2 = 162u^2v(3u + v)
162uv^3 + 54uv^3 = 216uv^3(3u + v)
Now, we have:
162u^2v(3u + v) + 216uv^3(3u + v)
We can see that both terms have a common factor of (3u + v), so we can factor it out:
(162u^2v + 216uv^3)(3u + v)
And that is the simplified form of the expression.
To get your result it would seem to be a whole lot easier to simply remove a factor 3 from the second bracket and to move that into the first bracket.
However the result would then be
(162u2v+54uv2)(3u+v),
which is different from your result, and not much of an improvement on the original.
I think that they are looking for something different.
Notice that 18uv is a common factor of the two terms in the first bracket, remove that and see where that leads.
Hi Tiggsy and George.
Assuming that 3u + v neq 0, simplify (54u^2 v + 18uv^2)(9u + 3v).
In a standard Australian High School I taught that simplify means "Get rid of the brackets and then collect like terms."
I think that is all that is required.
Hi Melody.
I take your point regarding simplification, and if the numbers, in this question, were different, I might agree with you.
If the thing to be simplified was (53u2v+18uv2)(10u+3v)
for example, all that you can do is remove uv from the first bracket or/and get rid of the brackets and collect up like terms.
What we are given though is (54u2v+18uv2)(9u+3v),
and this permits
18uv(3u+v).3(3u+v)=54uv(3u+v)2.
My feeling is that's what they are looking for.