Certainly! Let's delve into solving this step-by-step.
### Part (a): Factorize [tex]\( x^2 + 5x + 6 \)[/tex]
To factorize the quadratic expression [tex]\( x^2 + 5x + 6 \)[/tex], we look for two numbers that multiply to the constant term (6) and add up to the coefficient of the linear term (5).
1. List the pairs of factors of 6:
[tex]\[
(1, 6), (2, 3), (-1, -6), (-2, -3)
\][/tex]
2. Identify the pair that sums to 5:
- Among these pairs, [tex]\((2, 3)\)[/tex] is the pair that satisfies this condition because [tex]\( 2 + 3 = 5 \)[/tex].
3. Write the quadratic expression as a product of two binomials:
[tex]\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\][/tex]
Therefore, the factorized form of [tex]\( x^2 + 5x + 6 \)[/tex] is:
[tex]\[
(x + 2)(x + 3)
\][/tex]
### Part (b): Solve [tex]\( x^2 + 5x + 6 = 0 \)[/tex]
To solve the quadratic equation [tex]\( x^2 + 5x + 6 = 0 \)[/tex], we use the factored form obtained in Part (a):
1. Set the factored form equal to zero:
[tex]\[
(x + 2)(x + 3) = 0
\][/tex]
2. Apply the zero-product property, which states that if a product of factors is zero, at least one of the factors must be zero. Hence, solve for [tex]\( x \)[/tex] in each factor:
[tex]\[
x + 2 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
3. Solve each linear equation:
[tex]\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\][/tex]
[tex]\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\][/tex]
Therefore, the solutions to the equation [tex]\( x^2 + 5x + 6 = 0 \)[/tex] are:
[tex]\[
x = -3 \quad \text{and} \quad x = -2
\][/tex]
To summarize:
- Part (a): The factorization of [tex]\( x^2 + 5x + 6 \)[/tex] is [tex]\((x + 2)(x + 3)\)[/tex].
- Part (b): The solutions to the equation [tex]\( x^2 + 5x + 6 = 0 \)[/tex] are [tex]\( x = -3 \)[/tex] and [tex]\( x = -2 \)[/tex].
