Question

10 find the sum
\\(\frac{d^2 + d - 30}{d^2 + 3d - 40} + \frac{d^2 + 14d + 48}{d^2 - 2d - 48}\\)
\\(\boldsymbol{\text{a}}\\) \\(\frac{2d^2 + 15d + 18}{(d + 8)(d - 8)}\\)
\\(\boldsymbol{\text{b}}\\) \\(\frac{2d^2 + 15d + 18}{2d^2 + d - 88}\\)
\\(\boldsymbol{\text{c}}\\) \\(\frac{d^2 + 14d + 16}{(d + 8)(d - 8)}\\)
\\(\boldsymbol{\text{d}}\\) \\(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\\)

Explanation:

Step1: Factor each quadratic

Factor \(d^2 + d - 30\): Find two numbers that multiply to \(-30\) and add to \(1\). They are \(6\) and \(-5\), so \(d^2 + d - 30=(d + 6)(d - 5)\).
Factor \(d^2 + 3d - 40\): Numbers multiply to \(-40\), add to \(3\): \(8\) and \(-5\), so \(d^2 + 3d - 40=(d + 8)(d - 5)\).
Factor \(d^2 + 14d + 48\): Numbers multiply to \(48\), add to \(14\): \(6\) and \(8\), so \(d^2 + 14d + 48=(d + 6)(d + 8)\).
Factor \(d^2 - 2d - 48\): Numbers multiply to \(-48\), add to \(-2\): \(6\) and \(-8\), so \(d^2 - 2d - 48=(d + 6)(d - 8)\).

Step2: Substitute factored forms

The expression becomes:
\(\frac{(d + 6)(d - 5)}{(d + 8)(d - 5)} + \frac{(d + 6)(d + 8)}{(d + 6)(d - 8)}\)

Step3: Simplify each fraction

Simplify first fraction: Cancel \((d - 5)\) (assuming \(d eq5\)): \(\frac{d + 6}{d + 8}\).
Simplify second fraction: Cancel \((d + 6)\) (assuming \(d eq - 6\)): \(\frac{d + 8}{d - 8}\).

Step4: Find a common denominator

Common denominator is \((d + 8)(d - 8)\).
Rewrite fractions:
\(\frac{(d + 6)(d - 8)}{(d + 8)(d - 8)} + \frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\)

Step5: Expand numerators

Expand \((d + 6)(d - 8)\): \(d^2 - 8d + 6d - 48 = d^2 - 2d - 48\).
Expand \((d + 8)(d + 8)\): \(d^2 + 16d + 64\).

Step6: Add numerators

\(d^2 - 2d - 48 + d^2 + 16d + 64 = 2d^2 + 14d + 16\)? Wait, no—wait, let's re - check. Wait, first fraction after simplifying is \(\frac{d + 6}{d + 8}\), second is \(\frac{d + 8}{d - 8}\). Wait, maybe I made a mistake in step 4. Wait, correct common denominator: \((d + 8)(d - 8)\). So:
\(\frac{(d + 6)(d - 8)}{(d + 8)(d - 8)}+\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\)
Wait, no, first fraction: \(\frac{d + 6}{d + 8}=\frac{(d + 6)(d - 8)}{(d + 8)(d - 8)}\), second fraction: \(\frac{d + 8}{d - 8}=\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\)? No, wait, \(\frac{d + 8}{d - 8}=\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\) is wrong. Wait, \(\frac{d + 8}{d - 8}=\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\) is incorrect. The correct rewrite is \(\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\) is wrong. Wait, no: \(\frac{d + 8}{d - 8}=\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}\) is a mistake. Let's do it again.

Wait, first term: \(\frac{d + 6}{d + 8}\), second term: \(\frac{d + 8}{d - 8}\). To get common denominator \((d + 8)(d - 8)\), multiply first term by \(\frac{d - 8}{d - 8}\) and second term by \(\frac{d + 8}{d + 8}\).

So first term: \(\frac{(d + 6)(d - 8)}{(d + 8)(d - 8)}=\frac{d^2 - 8d + 6d - 48}{(d + 8)(d - 8)}=\frac{d^2 - 2d - 48}{(d + 8)(d - 8)}\).

Second term: \(\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)}=\frac{d^2 + 16d + 64}{(d + 8)(d - 8)}\).

Now add numerators: \(d^2 - 2d - 48 + d^2 + 16d + 64 = 2d^2 + 14d + 16\)? Wait, but let's check the answer options. Wait, maybe I made a mistake in factoring. Wait, let's re - factor the original quadratics.

Wait, original first numerator: \(d^2 + d - 30\). Let's check: \(d^2 + d - 30=(d + 6)(d - 5)\) (since \(6\times(-5)=-30\), \(6 + (-5)=1\))—correct. First denominator: \(d^2 + 3d - 40=(d + 8)(d - 5)\) ( \(8\times(-5)=-40\), \(8+(-5)=3\))—correct. Second numerator: \(d^2 + 14d + 48=(d + 6)(d + 8)\) ( \(6\times8 = 48\), \(6 + 8=14\))—correct. Second denominator: \(d^2 - 2d - 48=(d + 6)(d - 8)\) ( \(6\times(-8)=-48\), \(6+(-8)=-2\))—correct.

Simplify first fraction: \(\frac{(d + 6)(d - 5)}{(d + 8)(d - 5)}=\frac{d + 6}{d + 8}\) ( \(d eq5\)). Second fraction: \(\frac{(d + 6)(d + 8)}{(d + 6)(d - 8)}=\frac{d + 8}{d - 8}\) ( \(d eq - 6\)).

Now, \(\frac{d + 6}{d + 8}+\frac{d + 8}{d - 8}\). Let's find common denominator \((d + 8)(d - 8)\).

\(\frac{(d + 6)(d - 8)+(d + 8)(d + 8)}{(d + 8)(d - 8)}\).

Expand \((d + 6)(d - 8)=d^2 - 8d + 6d - 48=d^2 - 2d - 48\).

Expand \((d + 8)(d + 8)=d^2 + 16d + 64\).

Add: \(d^2 - 2d - 48 + d^2 + 16d + 64 = 2d^2 + 14d + 16\). Wait, but option D is \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\). But let's check the answer options again. Wait, maybe I made a mistake in the problem. Wait, let's check the original problem again.

Wait, the original problem is \(\frac{d^2 + d - 30}{d^2 + 3d - 40}+\frac{d^2 + 14d + 48}{d^2 - 2d - 48}\).

Wait, another approach: Let's not simplify first, but find common denominator as \((d^2 + 3d - 40)(d^2 - 2d - 48)\). But that's more complicated. Wait, the first denominator is \((d + 8)(d - 5)\), second is \((d + 6)(d - 8)\). Wait, maybe my initial simplification was wrong. Wait, no—if we don't cancel first, let's see:

\(\frac{(d^2 + d - 30)(d^2 - 2d - 48)+(d^2 + 14d + 48)(d^2 + 3d - 40)}{(d^2 + 3d - 40)(d^2 - 2d - 48)}\).

But that's too long. Alternatively, let's plug in a value for \(d\), say \(d = 0\).

Original expression: \(\frac{0 + 0 - 30}{0 + 0 - 40}+\frac{0 + 0 + 48}{0 - 0 - 48}=\frac{-30}{-40}+\frac{48}{-48}=\frac{3}{4}-1=-\frac{1}{4}\).

Now check option D: \(\frac{2(0)^2 + 14(0)+16}{(0 + 8)(0 - 8)}=\frac{16}{-64}=-\frac{1}{4}\). Correct.

Option A: \(\frac{0 + 0 + 18}{(8)(-8)}=\frac{18}{-64}=-\frac{9}{32} eq-\frac{1}{4}\).

Option B: \(\frac{0 + 0 + 18}{0 + 0 - 88}=\frac{18}{-88}=-\frac{9}{44} eq-\frac{1}{4}\).

Option C: \(\frac{0 + 0 + 16}{(8)(-8)}=\frac{16}{-64}=-\frac{1}{4}\)? Wait, option C numerator is \(d^2 + 14d + 16\), at \(d = 0\) it's \(16\), same as D? Wait, no, wait option D is \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\), option C is \(\frac{d^2 + 14d + 16}{(d + 8)(d - 8)}\). Wait, at \(d = 0\), option C: \(\frac{16}{-64}=-\frac{1}{4}\), option D: \(\frac{16}{-64}=-\frac{1}{4}\). Wait, that's a problem. Wait, maybe my plug - in is wrong. Let's try \(d = 1\).

Original expression: \(\frac{1 + 1 - 30}{1 + 3 - 40}+\frac{1 + 14 + 48}{1 - 2 - 48}=\frac{-28}{-36}+\frac{63}{-49}=\frac{7}{9}-\frac{9}{7}=\frac{49 - 81}{63}=-\frac{32}{63}\).

Option D: \(\frac{2(1)+14 + 16}{(9)(-7)}=\frac{2 + 14 + 16}{-63}=\frac{32}{-63}=-\frac{32}{63}\). Correct.

Option C: \(\frac{1 + 14 + 16}{(9)(-7)}=\frac{31}{-63} eq-\frac{32}{63}\). Ah, so option D is correct.

Answer:

D. \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\)