Question

problems 5 - 8: factor each of the following completely. notice that there are no equal signs in the expressions, which means that they are not equations. your answers should be factors, not single numbers.
5.) (2x^2 - 8x)
6.) (3x^2 - 5x - 8)
7.) (2x^2 + 3x + 1)
8.) (-\frac{1}{2}x^2 + 6x - 10)
problems 9 - 18: solve each of the following equations. write any non - integer answers in exact and simplified form.
simplifying checklist:

• terms that can be combined are combined; terms that cannot be combined are not combined.
• there are no rounded - off decimals in any of my answers.
• there are no fractions or decimals inside any of the square roots in any of my answers.
• there are no square roots in the denominators in any of my answers.
• there are no square roots that can be simplified further in any of my answers.
• any complex - imaginary answers are written in the form (a + bi), (a - bi), or (apm bi).

9.) (2x^2 + x = 0)
10.) (x^2 - 25 = 0)
11.) (x^2 + 8x + 7 = 0)
12.) (x^2 + 4x + 5 = 0)
13.) (-2(x + 1)^2 = -32)
14.) (x^2 + 6x - 1 = 0)

Explanation:

Step1: Factor out the GCF

The given equation is \(2x^{2}+x = 0\). The greatest common factor (GCF) of \(2x^{2}\) and \(x\) is \(x\). Factoring out \(x\), we get \(x(2x + 1)=0\).

Step2: Apply the zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So, for \(x(2x + 1)=0\), we have two cases:

• Case 1: \(x=0\)
• Case 2: \(2x + 1=0\). Solving \(2x+1 = 0\) for \(x\), we subtract 1 from both sides: \(2x=-1\), and then divide both sides by 2: \(x =-\frac{1}{2}\)

Answer:

\(x = 0\) or \(x=-\frac{1}{2}\)