Question
What is the factorization of the polynomial below?
3x^2+33x+54
0(3x+ 2)(x+9)
0(x+3)(x+9)
0(X+2)(x+27)
03(x+2)(x+9)
3x^2+33x+54
0(3x+ 2)(x+9)
0(x+3)(x+9)
0(X+2)(x+27)
03(x+2)(x+9)
Answers
GPT-4o mini
To factor the polynomial \( 3x^2 + 33x + 54 \), we first look for a common factor.
1. The coefficients \( 3 \), \( 33 \), and \( 54 \) all have a common factor of \( 3 \). Let's factor that out:
\[
3(x^2 + 11x + 18)
\]
2. Now we need to factor the quadratic \( x^2 + 11x + 18 \). We look for two numbers that multiply to \( 18 \) (the constant term) and add up to \( 11 \) (the coefficient of \( x \)). The numbers \( 2 \) and \( 9 \) work because:
\[
2 \times 9 = 18 \quad \text{and} \quad 2 + 9 = 11
\]
3. Therefore, we can factor \( x^2 + 11x + 18 \) as \( (x + 2)(x + 9) \).
4. Putting it all together with the common factor we factored out, we have:
\[
3(x + 2)(x + 9)
\]
Thus, the factorization of the polynomial \( 3x^2 + 33x + 54 \) is:
\[
\boxed{3(x + 2)(x + 9)}
\]
1. The coefficients \( 3 \), \( 33 \), and \( 54 \) all have a common factor of \( 3 \). Let's factor that out:
\[
3(x^2 + 11x + 18)
\]
2. Now we need to factor the quadratic \( x^2 + 11x + 18 \). We look for two numbers that multiply to \( 18 \) (the constant term) and add up to \( 11 \) (the coefficient of \( x \)). The numbers \( 2 \) and \( 9 \) work because:
\[
2 \times 9 = 18 \quad \text{and} \quad 2 + 9 = 11
\]
3. Therefore, we can factor \( x^2 + 11x + 18 \) as \( (x + 2)(x + 9) \).
4. Putting it all together with the common factor we factored out, we have:
\[
3(x + 2)(x + 9)
\]
Thus, the factorization of the polynomial \( 3x^2 + 33x + 54 \) is:
\[
\boxed{3(x + 2)(x + 9)}
\]