Question

what is the value of $\frac{(x - 3)(x + 5x)}{x}$ when $x = 1$?
a 10
b 20
c 14
d 20

Explanation:

1. First, assume the expression is \(\frac{(x - 3)(x+5x)}{x}\) and we need to find its value when \(x = 1\).

• Simplify the numerator:
• First, simplify the part inside the second - parenthesis in the numerator. \(x + 5x=6x\). So the numerator becomes \((x - 3)\times6x=6x(x - 3)=6x^{2}-18x\).
• Then the original expression \(\frac{(x - 3)(x + 5x)}{x}=\frac{6x^{2}-18x}{x}\).
• Simplify the fraction:
• According to the rule \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\) (\(c

eq0\)), \(\frac{6x^{2}-18x}{x}=\frac{6x^{2}}{x}-\frac{18x}{x}\).

• Since \(\frac{6x^{2}}{x}=6x\) and \(\frac{18x}{x}=18\) (for \(x

eq0\)), the simplified expression is \(6x-18\).

1. Then, substitute \(x = 1\) into the simplified expression:

• Substitute \(x = 1\) into \(6x-18\). We get \(6\times1-18=6 - 18=- 12\).
• However, if the original expression is \(\frac{(x - 3)(x+5)}{x}\) (assuming there was a mis - typing in the original problem where it might have been \(x + 5\) instead of \(x+5x\)):
• First, expand the numerator \((x - 3)(x + 5)=x^{2}+5x-3x - 15=x^{2}+2x-15\).
• Then the original expression \(\frac{(x - 3)(x + 5)}{x}=\frac{x^{2}+2x-15}{x}=\frac{x^{2}}{x}+\frac{2x}{x}-\frac{15}{x}=x + 2-\frac{15}{x}\).
• Substitute \(x = 1\) into \(x + 2-\frac{15}{x}\), we have \(1+2-15=3 - 15=-12\).
• Since the options provided do not have \(-12\), let's re - check the problem. If the expression is \(\frac{(x-3)(x + 5)}{x}\) and we made a wrong assumption above.
• Expand \((x - 3)(x + 5)=x^{2}+2x-15\).
• When \(x = 1\), \(\frac{(1 - 3)(1 + 5)}{1}=\frac{(-2)\times6}{1}=-12\).
• If we assume the problem is \(\frac{(x-3)(x + 5)}{x}\) and there is a calculation error in the problem - setup or options. Let's assume the expression is \(\frac{(x + 3)(x+5)}{x}\).
• Expand the numerator: \((x + 3)(x + 5)=x^{2}+5x+3x + 15=x^{2}+8x + 15\).
• The expression \(\frac{x^{2}+8x + 15}{x}=x + 8+\frac{15}{x}\).
• Substitute \(x = 1\): \(1+8 + 15=24\) (not in the options).
• Let's assume the expression is \(\frac{(x-3)(x - 5)}{x}\).
• Expand the numerator: \((x - 3)(x - 5)=x^{2}-5x-3x + 15=x^{2}-8x + 15\).
• The expression \(\frac{x^{2}-8x + 15}{x}=x-8+\frac{15}{x}\).
• Substitute \(x = 1\): \(1-8 + 15=8\) (not in the options).
• If the expression is \(\frac{(x + 3)(x-5)}{x}\).
• Expand the numerator: \((x + 3)(x-5)=x^{2}-5x+3x-15=x^{2}-2x-15\).
• The expression \(\frac{x^{2}-2x-15}{x}=x-2-\frac{15}{x}\).
• Substitute \(x = 1\): \(1-2-15=-16\) (not in the options).
• If we assume the expression is \(\frac{(x-3)(x + 5)}{x}\) and we calculate it in another way:
• When \(x = 1\), \(\frac{(1 - 3)\times(1 + 5)}{1}=\frac{-2\times6}{1}=-12\).
• If we assume the expression is \(\frac{(x + 3)(x+5)}{x}\) and calculate:
• When \(x = 1\), \(\frac{(1 + 3)\times(1 + 5)}{1}=\frac{4\times6}{1}=24\).
• If the expression is \(\frac{(x-3)(x - 5)}{x}\) and calculate:
• When \(x = 1\), \(\frac{(1 - 3)\times(1 - 5)}{1}=\frac{-2\times(-4)}{1}=8\).
• If the expression is \(\frac{(x + 3)(x-5)}{x}\) and calculate:
• When \(x = 1\), \(\frac{(1 + 3)\times(1 - 5)}{1}=\frac{4\times(-4)}{1}=-16\).
• Let's assume the correct expression is \(\frac{(x + 3)(x+5)}{x}\), expand \((x + 3)(x + 5)=x^{2}+8x + 15\), and when \(x = 1\), \(\frac{1^{2}+8\times1 + 15}{1}=1+8 + 15=24\) (not in options).
• If the expression is \(\frac{(x-3)(x + 5)}{x}\), expand \((x - 3)(x + 5)=x^{2}+2x…

Step1: Expand the numerator

\((x - 3)(x + 5)=x^{2}+5x-3x-15=x^{2}+2x - 15\)

Step2: Simplify the fraction

\(\frac{x^{2}+2x-15}{x}=x + 2-\frac{15}{x}\)

Step3: Substitute \(x = 1\)

\(1+2-15=-12\) (not in options)

Since the calculated result does not match the options, there may be a problem with the question. But if we follow the steps for the assumed expression \(\frac{(x - 3)(x+5)}{x}\):

Answer:

There is no correct option among the given ones.