Sure, let’s work through this problem step by step.
### Part a) Factorizing the Expression [tex]\(2x^2 + 11x + 5\)[/tex]
We need to factorize the quadratic expression [tex]\(2x^2 + 11x + 5\)[/tex]. We will find two binomials that, when multiplied, give us the original quadratic expression.
Given:
[tex]\[
2x^2 + 11x + 5 = (2x + \square)(x + \square)
\][/tex]
To determine the values that fill in these squares, consider the factored form.
From the result, we know the factors are:
[tex]\[
(2x + 1)(x + 5)
\][/tex]
So, filling in the squares, we get:
[tex]\[
2x^2 + 11x + 5 = (2x + 1)(x + 5)
\][/tex]
### Part b) Solving the Equation [tex]\(2x^2 + 11x + 5 = 0\)[/tex]
Now that we have the factorized form from part a), we can use it to solve the equation:
[tex]\[
2x^2 + 11x + 5 = 0
\][/tex]
Substituting the factored form, we get:
[tex]\[
(2x + 1)(x + 5) = 0
\][/tex]
To find the values of [tex]\(x\)[/tex], we set each factor equal to zero and solve:
1. For [tex]\(2x + 1 = 0\)[/tex]:
[tex]\[
2x + 1 = 0
\][/tex]
[tex]\[
2x = -1
\][/tex]
[tex]\[
x = -\frac{1}{2}
\][/tex]
2. For [tex]\(x + 5 = 0\)[/tex]:
[tex]\[
x + 5 = 0
\][/tex]
[tex]\[
x = -5
\][/tex]
Thus, the solutions to the quadratic equation [tex]\(2x^2 + 11x + 5 = 0\)[/tex] are:
[tex]\[
x = -5 \quad \text{and} \quad x = -\frac{1}{2}
\][/tex]
### Summary
a) The factorized form of the quadratic expression is:
[tex]\[
2x^2 + 11x + 5 = (2x + 1)(x + 5)
\][/tex]
b) The solutions to the equation [tex]\(2x^2 + 11x + 5 = 0\)[/tex] are:
[tex]\[
x = -5 \quad \text{and} \quad x = -\frac{1}{2}
\][/tex]
