QuestionEvaluate $\frac{x^{2}-x-20}{x-5}$ for $x=-1$ .

Studdy Solution

STEP 1

Assumptions1. We are given the function $\frac{x^{}-1 x-20}{x-5}$ . We are asked to evaluate this function at $x=-1$

STEP 2

We can simplify the function by factoring the numerator.
The numerator $x^{2}-1 x-20$ can be factored into $(x-5)(x+4)$ .

STEP 3

Substitute the factored form of the numerator back into the function.
$\frac{x^{2}-1 x-20}{x-5} = \frac{(x-5)(x+)}{x-5}$

STEP 4

Notice that the factor $(x-)$ appears in both the numerator and the denominator. We can cancel this factor out, as long as $x \neq$ .
$\frac{(x-)(x+4)}{x-} = x+4$

STEP 5

Now, we can substitute $x=-1$ into the simplified function $x+4$ .
$x+4 = -1+4$

STEP 6

Calculate the value of the function at $x=-1$ .
$-1+4 =3$ So, the value of the function $\frac{x^{2}-1 x-20}{x-5}$ at $x=-1$ is3.

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QuestionEvaluate $\frac{x^{2}-x-20}{x-5}$ for $x=-1$ .

Studdy Solution

STEP 1

Assumptions1. We are given the function $\frac{x^{}-1 x-20}{x-5}$ . We are asked to evaluate this function at $x=-1$

STEP 2

We can simplify the function by factoring the numerator.
The numerator $x^{2}-1 x-20$ can be factored into $(x-5)(x+4)$ .

STEP 3

Substitute the factored form of the numerator back into the function.
$\frac{x^{2}-1 x-20}{x-5} = \frac{(x-5)(x+)}{x-5}$

STEP 4

Notice that the factor $(x-)$ appears in both the numerator and the denominator. We can cancel this factor out, as long as $x \neq$ .
$\frac{(x-)(x+4)}{x-} = x+4$

STEP 5

Now, we can substitute $x=-1$ into the simplified function $x+4$ .
$x+4 = -1+4$

STEP 6

Calculate the value of the function at $x=-1$ .
$-1+4 =3$ So, the value of the function $\frac{x^{2}-1 x-20}{x-5}$ at $x=-1$ is3.

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QuestionEvaluate x2−x−20x−5\frac{x^{2}-x-20}{x-5}x−5x2−x−20​ for x=−1x=-1x=−1.

Studdy Solution

STEP 1

Assumptions1. We are given the function x−1x−20x−5\frac{x^{}-1 x-20}{x-5}x−5x−1x−20​ . We are asked to evaluate this function at x=−1x=-1x=−1

STEP 2

We can simplify the function by factoring the numerator.The numerator x2−1x−20x^{2}-1 x-20x2−1x−20 can be factored into (x−5)(x+4)(x-5)(x+4)(x−5)(x+4).

STEP 3

Substitute the factored form of the numerator back into the function.x2−1x−20x−5=(x−5)(x+)x−5\frac{x^{2}-1 x-20}{x-5} = \frac{(x-5)(x+)}{x-5}x−5x2−1x−20​=x−5(x−5)(x+)​

STEP 4

Notice that the factor (x−)(x-)(x−) appears in both the numerator and the denominator. We can cancel this factor out, as long as x≠x \neqx=.(x−)(x+4)x−=x+4\frac{(x-)(x+4)}{x-} = x+4x−(x−)(x+4)​=x+4

STEP 5

Now, we can substitute x=−1x=-1x=−1 into the simplified function x+4x+4x+4.x+4=−1+4x+4 = -1+4x+4=−1+4

STEP 6

Calculate the value of the function at x=−1x=-1x=−1.−1+4=3-1+4 =3−1+4=3So, the value of the function x2−1x−20x−5\frac{x^{2}-1 x-20}{x-5}x−5x2−1x−20​ at x=−1x=-1x=−1 is3.

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QuestionEvaluate $\frac{x^{2}-x-20}{x-5}$ for $x=-1$ .

Assumptions1. We are given the function $\frac{x^{}-1 x-20}{x-5}$ . We are asked to evaluate this function at $x=-1$

We can simplify the function by factoring the numerator.
The numerator $x^{2}-1 x-20$ can be factored into $(x-5)(x+4)$ .

Substitute the factored form of the numerator back into the function.
$\frac{x^{2}-1 x-20}{x-5} = \frac{(x-5)(x+)}{x-5}$

Notice that the factor $(x-)$ appears in both the numerator and the denominator. We can cancel this factor out, as long as $x \neq$ .
$\frac{(x-)(x+4)}{x-} = x+4$

Now, we can substitute $x=-1$ into the simplified function $x+4$ .
$x+4 = -1+4$

Calculate the value of the function at $x=-1$ .
$-1+4 =3$ So, the value of the function $\frac{x^{2}-1 x-20}{x-5}$ at $x=-1$ is3.