Question

1. solve the equation $2x^2 - x - 21 = 0$ by factoring. be sure to show each step and include your solutions in your answer.

enter your solutions and your work or explanations in the box provided.

Explanation:

Step1: Find two numbers

We need two numbers that multiply to \(2\times(-21)= -42\) and add up to \(-1\). The numbers are \(-7\) and \(6\) since \(-7\times6 = -42\) and \(-7 + 6=-1\).

Step2: Rewrite the middle term

Rewrite the equation \(2x^{2}-x - 21 = 0\) as \(2x^{2}+6x-7x - 21 = 0\).

Step3: Group and factor

Group the first two terms and the last two terms: \((2x^{2}+6x)-(7x + 21)=0\). Factor out the common factors from each group: \(2x(x + 3)-7(x + 3)=0\).

Step4: Factor out the common binomial

Factor out \((x + 3)\) from the expression: \((2x - 7)(x + 3)=0\).

Step5: Solve for x

Set each factor equal to zero:

• \(2x-7 = 0\) gives \(2x=7\) so \(x=\frac{7}{2}\)
• \(x + 3=0\) gives \(x=-3\)

Answer:

The solutions of the equation \(2x^{2}-x - 21 = 0\) are \(x = \frac{7}{2}\) and \(x=-3\)