Question

simplify the following polynomial expression and state restriction.

a)\frac{3}{x^2-7x+10}-\frac{5x}{x^2-4}

Answer 1

[AI-generated answer — Bot GPT 3.5]

To simplify the expression, we first find a common denominator for the two fractions. The common denominator would be the product of the two denominators: (x^2-7x+10)(x^2-4).

Next, we use this common denominator to express each fraction with the common denominator:

\frac{3}{x^2-7x+10} - \frac{5x}{x^2-4} = \frac{3(x^2-4)}{(x^2-7x+10)(x^2-4)} - \frac{5x(x^2-7x+10)}{(x^2-7x+10)(x^2-4)}

Simplifying further:

= \frac{3x^2-12}{(x^2-7x+10)(x^2-4)} - \frac{5x^3-35x^2+50x}{(x^2-7x+10)(x^2-4)}

Now, to combine the two fractions, we need a common numerator. We can find a common numerator by multiplying each term by the expansion of the other numerator. However, this would result in a very long expression.

Therefore, we can leave the expression as it is. The restriction would be any value of x for which the denominators (x^2-7x+10) and (x^2-4) are equal to zero. To find these values, we can set each denominator equal to zero and solve for x:

For x^2-7x+10 = 0:
(x-5)(x-2) = 0
x = 5 or x = 2

For x^2-4 = 0:
(x+2)(x-2) = 0
x = -2 or x = 2

Therefore, the restriction is x ≠ -2, 2, 5.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To simplify the given expression, we first need to find a common denominator for both fractions.

The denominators of the two fractions are \(x^2-7x+10\) and \(x^2-4\). To find a common denominator, we need to factor both denominators.

The first denominator can be factored as \((x-5)(x-2)\), and the second denominator can be factored as \((x-2)(x+2)\).

Now we have the fractions as:
\[\frac{3}{(x-5)(x-2)} - \frac{5x}{(x-2)(x+2)}\]

To combine these fractions, we need a common denominator, which is \((x-5)(x-2)(x+2)\).

Now we can rewrite the expression with the common denominator:
\[\frac{3(x+2)}{(x-5)(x-2)(x+2)} - \frac{5x(x-5)}{(x-5)(x-2)(x+2)}\]

Next, we can simplify the expression by combining the numerators:
\[\frac{3(x+2)-5x(x-5)}{(x-5)(x-2)(x+2)}\]

Expanding the terms in the numerator:
\[\frac{3x+6-5x^2+25x}{(x-5)(x-2)(x+2)}\]

Combining like terms in the numerator:
\[\frac{-5x^2+28x+6}{(x-5)(x-2)(x+2)}\]

The final simplified expression is:
\[\frac{-5x^2+28x+6}{(x-5)(x-2)(x+2)}\]

Restrictions:
Now, let's consider the restrictions. Since the original denominator includes \(x^2-7x+10\), we can find the restrictions by setting this factor equal to zero and solving for \(x\):
\[x^2-7x+10 = 0\]

This quadratic equation factors as \((x-5)(x-2) = 0\), which means that \(x = 5\) or \(x = 2\).

Therefore, the restrictions on the expression are \(x \neq 5\) and \(x \neq 2\).

Answer 3

[AI-generated answer — Explain Bot]

To simplify the given polynomial expression, we will first find the least common denominator (LCD) of the fractions. The LCD is the least common multiple of the denominators.

The first denominator is \(x^2-7x+10\), which can be factored into \((x-2)(x-5)\).

The second denominator is \(x^2-4\), which is a difference of squares and can be factored into \((x+2)(x-2)\).

The LCD is the product of both denominators, so the LCD is \((x-2)(x-5)(x+2)\).

Next, we will rewrite each fraction with the LCD as the denominator:

\(\frac{3}{x^2-7x+10} = \frac{3}{(x-2)(x-5)} \cdot \frac{(x+2)}{(x+2)} = \frac{3(x+2)}{(x-2)(x-5)(x+2)}\)

\(\frac{5x}{x^2-4} = \frac{5x}{(x+2)(x-2)} \cdot \frac{(x-5)}{(x-5)} = \frac{5x(x-5)}{(x-2)(x-5)(x+2)}\)

Now that both fractions have the same denominator, we can combine them:

\(\frac{3(x+2)}{(x-2)(x-5)(x+2)} - \frac{5x(x-5)}{(x-2)(x-5)(x+2)}\)

To simplify, we can combine the numerators:

\(\frac{3(x+2) - 5x(x-5)}{(x-2)(x-5)(x+2)}\)

Expanding the expressions in the numerator:

\(\frac{3x + 6 - 5x^2 + 25x}{(x-2)(x-5)(x+2)}\)

Combining like terms:

\(\frac{-5x^2 + 28x + 6}{(x-2)(x-5)(x+2)}\)

Therefore, the simplified polynomial expression is \(\frac{-5x^2 + 28x + 6}{(x-2)(x-5)(x+2)}\).

To find any restrictions, we need to look for values of \(x\) that make the denominators zero. In this case, we have \(x-2=0\), \(x-5=0\), and \(x+2=0\).

Solving these equations, we find that the restrictions are \(x=2\), \(x=5\), and \(x=-2\). These values of \(x\) should be excluded from the domain to ensure that the expression is defined.