Question

1. simplify the expression \\(\frac{p - 3}{p - 4} + \frac{2p^2 + 9p + 4}{p^2 - 9p + 20} \div \frac{p^2 + 3p - 4}{2p^2 - 9p - 5}\\). indicate any non - permissible values.

Explanation:

Step1: Factor all quadratics

First, factor each quadratic expression:
- \(p^2 - 9p + 20=(p - 4)(p - 5)\)
- \(2p^2+9p + 4=(2p + 1)(p + 4)\)
- \(p^2+3p - 4=(p + 4)(p - 1)\)
- \(2p^2-9p - 5=(2p + 1)(p - 5)\)

So the expression becomes:
\(\frac{p - 3}{p - 4}+\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\div\frac{(p + 4)(p - 1)}{(2p + 1)(p - 5)}\)

Step2: Change division to multiplication (reciprocal)

Recall that \(a\div b=a\times\frac{1}{b}\), so:
\(\frac{p - 3}{p - 4}+\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\times\frac{(2p + 1)(p - 5)}{(p + 4)(p - 1)}\)

Step3: Cancel common factors

Cancel out common factors in the multiplication part:
- \((p + 4)\) cancels, \((p - 5)\) cancels, \((2p + 1)\) cancels (note: \(2p+1 eq0\Rightarrow p eq-\frac{1}{2}\), \(p + 4 eq0\Rightarrow p eq - 4\), \(p - 5 eq0\Rightarrow p eq5\) from original denominators and factors)
After cancellation, the multiplication part becomes \(\frac{(2p + 1)}{(p - 4)}\times\frac{(2p + 1)}{(p - 1)}\)? Wait, no, wait:
Wait, \(\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\times\frac{(2p + 1)(p - 5)}{(p + 4)(p - 1)}=\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)? No, wait, \((2p + 1)\) is in numerator once and once? Wait, no: first numerator has \((2p + 1)\), second numerator has \((2p + 1)\), first denominator has \((p - 4)(p - 5)\), second denominator has \((p + 4)(p - 1)\), first numerator has \((p + 4)\), second numerator has \((p - 5)\). So:
\((p + 4)\) cancels, \((p - 5)\) cancels, one \((2p + 1)\) in numerator and one in numerator? Wait, no:
First fraction numerator: \((2p + 1)(p + 4)\), second fraction numerator: \((2p + 1)(p - 5)\)
First fraction denominator: \((p - 4)(p - 5)\), second fraction denominator: \((p + 4)(p - 1)\)
So when we multiply, it's \(\frac{(2p + 1)(p + 4)\times(2p + 1)(p - 5)}{(p - 4)(p - 5)\times(p + 4)(p - 1)}\)
Cancel \((p + 4)\), \((p - 5)\), so we get \(\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)? Wait, no, \((2p + 1)\) is multiplied twice? Wait, no, first numerator has \((2p + 1)\), second numerator has \((2p + 1)\), so numerator is \((2p + 1)(2p + 1)=(2p + 1)^2\), denominator is \((p - 4)(p - 1)\)

Wait, no, actually:
\(\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\times\frac{(2p + 1)(p - 5)}{(p + 4)(p - 1)}=\frac{(2p + 1)\cancel{(p + 4)}}{(p - 4)\cancel{(p - 5)}}\times\frac{(2p + 1)\cancel{(p - 5)}}{\cancel{(p + 4)}(p - 1)}=\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)

Now the expression is \(\frac{p - 3}{p - 4}+\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)

Step4: Find a common denominator

The common denominator is \((p - 4)(p - 1)\)
Rewrite \(\frac{p - 3}{p - 4}\) with denominator \((p - 4)(p - 1)\):
\(\frac{(p - 3)(p - 1)}{(p - 4)(p - 1)}\)

So now the expression is:
\(\frac{(p - 3)(p - 1)+(2p + 1)^2}{(p - 4)(p - 1)}\)

Step5: Expand numerator

Expand \((p - 3)(p - 1)=p^2 - p - 3p + 3=p^2 - 4p + 3\)
Expand \((2p + 1)^2 = 4p^2 + 4p + 1\)
Add them together:
\(p^2 - 4p + 3+4p^2 + 4p + 1 = 5p^2 + 4\)? Wait, no:
Wait, \(p^2 - 4p + 3+4p^2 + 4p + 1=(p^2 + 4p^2)+(-4p + 4p)+(3 + 1)=5p^2 + 4\)? Wait, that can't be right. Wait, no:
Wait, \((p - 3)(p - 1)=p^2 - p - 3p + 3=p^2 - 4p + 3\)
\((2p + 1)^2=4p^2 + 4p + 1\)
Adding: \(p^2 - 4p + 3+4p^2 + 4p + 1 = 5p^2 + 4\)? Wait, the \(p\) terms cancel: \(-4p + 4p = 0\), so \(p^2+4p^2=5p^2\), \(3 + 1 = 4\). So numerator is \(5p^2 + 4\)? Wait, that seems odd. Wait, maybe I made a mistake in the multiplication step.

Wait, let's go back to the division step. Maybe I messed up the reciprocal.

Wait, original division: \(\frac{2p^2 + 9p + 4}{p^2 - 9p + 20}\div\frac{p^2 + 3p - 4}{2p^2 - 9p - 5}\)

Which is \(\frac{2p^2 + 9p + 4}{p^2 - 9p + 20}\times\frac{2p^2 - 9p - 5}{p^2 + 3p - 4}\)

Factor each:

\(2p^2 + 9p + 4=(2p + 1)(p + 4)\) (correct)

\(p^2 - 9p + 20=(p - 4)(p - 5)\) (correct)

\(2p^2 - 9p - 5=(2p + 1)(p - 5)\) (correct)

\(p^2 + 3p - 4=(p + 4)(p - 1)\) (correct)

So multiplying:

\(\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\times\frac{(2p + 1)(p - 5)}{(p + 4)(p - 1)}\)

Now, cancel \((p + 4)\) (so \(p eq - 4\)), \((p - 5)\) (so \(p eq5\)), and we have \(\frac{(2p + 1)(2p + 1)}{(p - 4)(p - 1)}=\frac{(2p + 1)^2}{(p - 4)(p - 1)}\) (correct)

Then the original expression is \(\frac{p - 3}{p - 4}+\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)

Common denominator: \((p - 4)(p - 1)\)

So \(\frac{(p - 3)(p - 1)}{(p - 4)(p - 1)}+\frac{(2p + 1)^2}{(p - 4)(p - 1)}\)

Now expand \((p - 3)(p - 1)=p^2 - p - 3p + 3=p^2 - 4p + 3\)

Expand \((2p + 1)^2=4p^2 + 4p + 1\)

Add them: \(p^2 - 4p + 3 + 4p^2 + 4p + 1 = 5p^2 + 4\)? Wait, that's correct? The \(p\) terms cancel. So numerator is \(5p^2 + 4\), denominator is \((p - 4)(p - 1)\)

Wait, but let's check again. Maybe I made a mistake in the problem. Wait, the original expression is \(\frac{p - 3}{p - 4}+\frac{2p^2 + 9p + 4}{p^2 - 9p + 20}\div\frac{p^2 + 3p - 4}{2p^2 - 9p - 5}\)

Alternatively, maybe I messed up the sign in \(2p^2 - 9p - 5\). Let's re - factor \(2p^2 - 9p - 5\):

\(2p^2 - 9p - 5\): multiply \(2\times(-5)=-10\), find two numbers that multiply to - 10 and add to - 9: - 10 and + 1. So \(2p^2 - 10p + p - 5 = 2p(p - 5)+1(p - 5)=(2p + 1)(p - 5)\). Correct.

\(p^2 + 3p - 4\): multiply \(1\times(-4)=-4\), numbers: 4 and - 1. So \(p^2 + 4p - p - 4=p(p + 4)-1(p + 4)=(p + 4)(p - 1)\). Correct.

\(2p^2 + 9p + 4\): multiply \(2\times4 = 8\), numbers: 8 and 1. So \(2p^2+8p + p + 4 = 2p(p + 4)+1(p + 4)=(2p + 1)(p + 4)\). Correct.

\(p^2 - 9p + 20\): multiply \(1\times20 = 20\), numbers: - 4 and - 5. So \(p^2-4p - 5p + 20=p(p - 4)-5(p - 4)=(p - 4)(p - 5)\). Correct.

So the expansion is correct.

Now, non - permissible values:

We need to find values of \(p\) that make any denominator zero.

From the original denominators and the factored forms:

1. \(p - 4 = 0\Rightarrow p = 4\)
2. \(p^2 - 9p + 20=(p - 4)(p - 5)=0\Rightarrow p = 4\) or \(p = 5\)
3. \(p^2+3p - 4=(p + 4)(p - 1)=0\Rightarrow p=-4\) or \(p = 1\)
4. \(2p^2 - 9p - 5=(2p + 1)(p - 5)=0\Rightarrow p=-\frac{1}{2}\) or \(p = 5\)

Also, from the canceled factors:

- \(p + 4 = 0\Rightarrow p=-4\) (already considered)
- \(p - 5 = 0\Rightarrow p = 5\) (already considered)
- \(2p+1 = 0\Rightarrow p=-\frac{1}{2}\) (already considered)
- \(p - 1 = 0\Rightarrow p = 1\) (already considered)

So the non - permissible values are \(p=-4,p = 1,p = 4,p = 5,p=-\frac{1}{2}\)

Now, simplify the numerator:

Wait, maybe I made a mistake in the addition. Let's re - expand \((p - 3)(p - 1)\) and \((2p + 1)^2\):

\((p - 3)(p - 1)=p^2 - p - 3p + 3=p^2 - 4p + 3\)

\((2p + 1)^2=4p^2 + 4p + 1\)

Adding them: \(p^2 - 4p + 3+4p^2 + 4p + 1=5p^2 + 4\)

So the simplified expression is \(\frac{5p^2 + 4}{(p - 4)(p - 1)}\) with non - permissible values \(p=-4,p = 1,p = 4,p = 5,p=-\frac{1}{2}\)

Wait, but let's check with another approach. Maybe I messed up the division step. Let's do the division again:

\(\frac{2p^2 + 9p + 4}{p^2 - 9p + 20}\div\frac{p^2 + 3p - 4}{2p^2 - 9p - 5}=\frac{2p^2 + 9p + 4}{p^2 - 9p + 20}\times\frac{2p^2 - 9p - 5}{p^2 + 3p - 4}\)

After canceling, we have \(\frac{(2p + 1)}{(p - 4)}\times\frac{(2p + 1)}{(p - 1)}\)? Wait, no, \(\frac{(2p + 1)(p + 4)}{(p - 4)(p - 5)}\times\frac{(2p + 1)(p - 5)}{(p + 4)(p - 1)}=\frac{(2p + 1)\times(2p + 1)}{(p - 4)\times(p - 1)}=\frac{(2p + 1)^2}{(p - 4)(p - 1)}\) (correct)

Then adding \(\frac{p - 3}{p - 4}=\frac{(p - 3)(p - 1)}{(p - 4)(p - 1)}\)

So \(\frac{(p - 3)(p - 1)+(2p + 1)^2}{(p - 4)(p - 1)}=\frac{p^2 - 4p + 3+4p^2 + 4p + 1}{(p - 4)(p - 1)}=\frac{5p^2 + 4}{(p - 4)(p - 1)}\)

Yes, that's correct.

Answer:

The simplified expression is \(\frac{5p^{2}+4}{(p - 4)(p - 1)}\) and the non - permissible values are \(p=-4,p = 1,p = 4,p = 5,p=-\frac{1}{2}\)