Question 191276
Says to factor by grouping? 
40a^4-120a^2+90

<pre><font size = 4 color = "indigo"><b>
{{{40a^4-120a^2+90}}}

First factor out {{{10}}}

{{{10(4a^4-12a^2+9)}}}

In the parentheses, multiply the first coefficient 4
by the last coefficient 9, getting 36.  Write down
all pairs of integer factors whose product is 36.
They are

1*36
2*18
3*12
4*9
6*6

Since the last sign in the parentheses is +, we
add those:

1+36 = 37
2+18 = 20
3+12 = 15
 4+9 = 13 
 6+6 = 12

Since the middle coefficient of {{{(4a^4-12a^2+9)}}}
is 12, and since 12 = 6+6 we write -12a^2 as -6a^2-6a^2

{{{10(4a^4-12a^2+9)}}}

{{{10(4a^4-6a^2-6a^2+9)}}}

Factor the first two terms in the parentheses by factoring 
out {{{2a^2}}}

{{{10(2a^2(2a^2-3)-6a^2+9)}}}
 
Factor the last two terms in the parentheses by factoring 
out {{{-3}}}

{{{10(2a^2(2a^2-3)-3(2a^2-3))}}}

Inside the outer parentheses take out 
the inner parentheses {{{(2a^2-3)}}}

{{{10(2a^2-3)(2a^2-3)}}}

Notice the parenthesis can be written once as squared:

{{{10(2a^2-3)^2}}}

Edwin</pre>