Question

6.4 trinomials where a does not equal 1: p(1 point)factor:2x² - x - 1 =you have attempted this problem 0 times.you have unlimited attempts remaining.

Explanation:

Step1: Multiply the coefficient of \(x^{2}\) and the constant term

For the trinomial \(2x^{2}-x - 1\), \(a = 2\), \(b=-1\), \(c=-1\). Calculate \(a\times c=2\times(-1)=-2\)

Step2: Find two numbers that multiply to \(a\times c\) and add up to \(b\)

We need two numbers \(m\) and \(n\) such that \(m\times n=-2\) and \(m + n=-1\). The numbers are \(m = 1\) and \(n=-2\) since \(1\times(-2)=-2\) and \(1+(-2)=-1\)

Step3: Rewrite the middle term

Rewrite \(-x\) as \(x-2x\). So \(2x^{2}-x - 1=2x^{2}+x-2x - 1\)

Step4: Group the terms

Group the terms as \((2x^{2}+x)-(2x + 1)\)

Step5: Factor out the common factors from each group

Factor out \(x\) from the first group and \(-1\) from the second group: \(x(2x + 1)-1(2x + 1)\)

Step6: Factor out the common binomial factor

Factor out \((2x + 1)\), we get \((2x + 1)(x-1)\)

Answer:

\((2x + 1)(x - 1)\)