Question
or
Factor completely.
2n2+21n+10
Answers
GPT-4o mini
To factor the quadratic expression \(2n^2 + 21n + 10\), we can look for two numbers that multiply to the product of the leading coefficient (2) and the constant term (10), which is \(2 \cdot 10 = 20\), and add up to the middle coefficient (21).
The two numbers that meet these criteria are 20 and 1, since:
- \(20 \times 1 = 20\)
- \(20 + 1 = 21\)
Next, we can rewrite the middle term (21n) using these two numbers:
\[
2n^2 + 20n + 1n + 10
\]
Now, we group the terms:
\[
(2n^2 + 20n) + (1n + 10)
\]
Factoring out the common factors in each group:
\[
2n(n + 10) + 1(n + 10)
\]
Now, we can factor out the common binomial \(n + 10\):
\[
(2n + 1)(n + 10)
\]
Thus, the completely factored form of \(2n^2 + 21n + 10\) is:
\[
(2n + 1)(n + 10)
\]
The two numbers that meet these criteria are 20 and 1, since:
- \(20 \times 1 = 20\)
- \(20 + 1 = 21\)
Next, we can rewrite the middle term (21n) using these two numbers:
\[
2n^2 + 20n + 1n + 10
\]
Now, we group the terms:
\[
(2n^2 + 20n) + (1n + 10)
\]
Factoring out the common factors in each group:
\[
2n(n + 10) + 1(n + 10)
\]
Now, we can factor out the common binomial \(n + 10\):
\[
(2n + 1)(n + 10)
\]
Thus, the completely factored form of \(2n^2 + 21n + 10\) is:
\[
(2n + 1)(n + 10)
\]