Question

select all of the rational roots of the following polynomial equation.\\(2x^2 + 3x + 1 = 0\\)\\(\square -\frac{1}{2}\\)\\(\square \frac{1}{2}\\)\\(\square -1\\)\\(\square 1\\)

Explanation:

Step1: Factor the quadratic equation

We factor \(2x^{2}+3x + 1\). We need two numbers that multiply to \(2\times1 = 2\) and add up to \(3\). The numbers are \(1\) and \(2\). So we rewrite the middle term:
\(2x^{2}+2x+x + 1=0\)
Group the terms:
\((2x^{2}+2x)+(x + 1)=0\)
Factor out the common factors from each group:
\(2x(x + 1)+1(x + 1)=0\)
Then we have \((2x + 1)(x + 1)=0\)

Step2: Solve for x

Set each factor equal to zero:

• For \(2x+1 = 0\), we solve for \(x\):

\(2x=-1\)
\(x=-\frac{1}{2}\)

• For \(x + 1=0\), we solve for \(x\):

\(x=-1\)

Answer:

\(-\frac{1}{2}\), \(-1\) (corresponding to the options: \(\boldsymbol{-\frac{1}{2}}\) and \(\boldsymbol{-1}\))