Question

In this activity, you’ll use the inspection method to rewrite a rational expression, a(x)/b(x), in the form q(x) + r(x)/b(x).Answer these questions to step through the process of rewriting x^2-5x+7/x-9Part ACan the polynomial in the numerator of the expression x^2-5x+7/x-9 be factored to derive (x-9) as a factor?Answer is noPart DWhat number must be added to the numerator to get the new constant term you identified in Part C?Part EAdd the number you calculated in part D to the numerator, and then subtract the number to keep the value of the expression unchanged.Part F Rewrite the numerator so it contains a trinomial that can be faced with x-9 as a common factor, and then write it in the factored formPart GRewrite the expression you found in part F as the sum of two rational expressions with (x-9) as their common denominator Part HReduce the first fraction and write the expression in this format:A(x)/b(x) = q(x)+ r(x)/b(x)

Answer 1

The expression is:

`(x^2-5x+7)/(x-9)`

Part B

To get -9 to -5, we need to add 4. This is important because the factored form will be something like this:

`x^2-5x+7=(x-9)(x+a)`

And when we distribute it, the middle term will be the sum of -9 and a, so we if we want it to be -5 (as the given expression) a has to be 4.

Part C

Now, looking to the constant part, it will be the multiplication of -9 and a, since we know that a is 4, the constant term is:

`-9\cdot4=-36`

So, we need a constant term of -36 in the numerator.

Part D

Since we already got 7 in the numerator, we have to add -43 to get it to -36.

Part E

`(x^2-5x+7)/(x-9)=(x^2-5x+7+(-43)-(-43))/(x-9)=(x^2-5x-36+43)/(x-9)`

Part F

`(x^2-5x+-36+43)/(x-9)=((x-9)(x+4)+43)/(x-9)`

Part G

`((x-9)(x+4)+43)/(x-9)=((x-9)(x+4))/(x-9)+(43)/(x-9)`

Part H

`((x-9)(x+4))/(x-9)+(43)/(x-9)=x+4+(43)/(x-9)`

So:

`(x^2-5x+7)/(x-9)=x+4+(43)/(x-9)`