Question
Factor the expression completely.
12n2+51n+12
please?
12n2+51n+12
please?
Answers
GPT-4o mini
To factor the expression \( 12n^2 + 51n + 12 \) completely, we can use the method of factoring by grouping.
First, we will find two numbers that multiply to \( 12 \times 12 = 144 \) (the product of the coefficient of \( n^2 \) and the constant term) and add up to \( 51 \) (the coefficient of \( n \)).
The two numbers that meet these conditions are \( 48 \) and \( 3 \) because:
- \( 48 \times 3 = 144 \)
- \( 48 + 3 = 51 \)
Next, we can rewrite the middle term \( 51n \) using \( 48n \) and \( 3n \):
\[
12n^2 + 48n + 3n + 12
\]
Now we can group the terms:
\[
(12n^2 + 48n) + (3n + 12)
\]
Factor out the common factors from each group:
\[
12n(n + 4) + 3(n + 4)
\]
Now we can factor out the common binomial factor \( (n + 4) \):
\[
(n + 4)(12n + 3)
\]
The expression \( 12n + 3 \) can be further factored out:
\[
3(4n + 1)
\]
Thus, the completely factored form of the original expression is:
\[
3(n + 4)(4n + 1)
\]
So the final answer is:
\[
3(n + 4)(4n + 1)
\]
First, we will find two numbers that multiply to \( 12 \times 12 = 144 \) (the product of the coefficient of \( n^2 \) and the constant term) and add up to \( 51 \) (the coefficient of \( n \)).
The two numbers that meet these conditions are \( 48 \) and \( 3 \) because:
- \( 48 \times 3 = 144 \)
- \( 48 + 3 = 51 \)
Next, we can rewrite the middle term \( 51n \) using \( 48n \) and \( 3n \):
\[
12n^2 + 48n + 3n + 12
\]
Now we can group the terms:
\[
(12n^2 + 48n) + (3n + 12)
\]
Factor out the common factors from each group:
\[
12n(n + 4) + 3(n + 4)
\]
Now we can factor out the common binomial factor \( (n + 4) \):
\[
(n + 4)(12n + 3)
\]
The expression \( 12n + 3 \) can be further factored out:
\[
3(4n + 1)
\]
Thus, the completely factored form of the original expression is:
\[
3(n + 4)(4n + 1)
\]
So the final answer is:
\[
3(n + 4)(4n + 1)
\]