2pi(x^2 + 4x + 4) + 2pi(x^2 + 7x + 10)
How do I write this as a polynomial in standard form (by factoring out 2pi)?
With 2π factored out…
(x^2 + 4x + 4) + (x^2 + 7x + 10) =
2x^2 + 11x + 14 or
2π(2x^2 + 11x + 14)
I don't understand how you did 2x^2.
ummm, that would be x^2 + x^2 = 2x^2
just as 4x+7x = 11x
and 4+10 = 14
just collect all the terms with the same power x^2 means 1x^2, so x^2+x^2 = 1x^2 + 1x^2 = 2x^2
To write the given expression as a polynomial in standard form by factoring out 2pi, we need to distribute the 2pi to each term within the parentheses. Let's go step by step:
2pi(x^2 + 4x + 4) + 2pi(x^2 + 7x + 10)
First, let's factor out 2pi from the first parentheses:
2pi * (x^2 + 4x + 4)
To factor out 2pi, we multiply each term within the parentheses by 2pi:
= 2pi * x^2 + 2pi * 4x + 2pi * 4
Simplifying, we get:
= 2pi * x^2 + 8pi * x + 8pi
Now, let's do the same for the second parentheses:
2pi * (x^2 + 7x + 10)
= 2pi * x^2 + 2pi * 7x + 2pi * 10
= 2pi * x^2 + 14pi * x + 20pi
Now, we can combine the two expressions:
(2pi * x^2 + 8pi * x + 8pi) + (2pi * x^2 + 14pi * x + 20pi)
Grouping common terms, we have:
(2pi * x^2 + 2pi * x^2) + (8pi * x + 14pi * x) + (8pi + 20pi)
= 4pi * x^2 + 22pi * x + 28pi
Therefore, the given expression, when factored out with 2pi, can be written as the polynomial 4pi * x^2 + 22pi * x + 28pi in standard form.