Question

20 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}{2d^2 + d - 88} \\) \\( \boldsymbol{\text{b}} \\) \\( \frac{2d^2 + 14d + 16}{(d + 8)(d - 8)} \\) \\( \boldsymbol{\text{c}} \\) \\( \frac{2d^2 + 15d + 18}{(d + 8)(d - 8)} \\) \\( \boldsymbol{\text{d}} \\) \\( \frac{d^2 + 14d + 16}{(d + 8)(d - 8)} \\)

Explanation:

Step1: Factor numerators and denominators

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

Substitute these factorizations into the original expression:
\[ \frac{(d + 6)(d - 5)}{(d + 8)(d - 5)}+\frac{(d + 6)(d + 8)}{(d - 8)(d + 6)} \]

Step2: Simplify each fraction

For the first fraction, cancel \((d - 5)\) (assuming \(d eq5\)): \(\frac{(d + 6)(d - 5)}{(d + 8)(d - 5)}=\frac{d + 6}{d + 8}\).
For the second fraction, cancel \((d + 6)\) (assuming \(d eq - 6\)): \(\frac{(d + 6)(d + 8)}{(d - 8)(d + 6)}=\frac{d + 8}{d - 8}\).

Now the expression becomes \(\frac{d + 6}{d + 8}+\frac{d + 8}{d - 8}\).

Step3: Find a common denominator and add

The common denominator is \((d + 8)(d - 8)\).
Rewrite each fraction with the common denominator:
\[ \frac{(d + 6)(d - 8)}{(d + 8)(d - 8)}+\frac{(d + 8)(d + 8)}{(d + 8)(d - 8)} \]

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

Add the numerators:
\[ \frac{(d^2 - 2d - 48)+(d^2 + 16d + 64)}{(d + 8)(d - 8)}=\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)} \] Wait, no, wait: Wait, \((d + 6)(d - 8)=d^2 - 8d + 6d - 48=d^2 - 2d - 48\)? Wait, no, \(6\times(-8)=-48\), \(d\times(-8)=-8d\), \(6\times d = 6d\), so \(-8d+6d=-2d\), correct. Then \((d + 8)(d + 8)=d^2 + 16d + 64\). Then adding numerators: \(d^2 - 2d - 48 + d^2 + 16d + 64 = 2d^2 + 14d + 16\)? Wait, no, \(-2d + 16d = 14d\), \(-48 + 64 = 16\), yes. Wait, but let's check again. Wait, maybe I made a mistake in factoring? Wait, no, let's re - do the addition. Wait, original fractions after simplification: \(\frac{d + 6}{d + 8}+\frac{d + 8}{d - 8}\). Let's compute numerator: \((d + 6)(d - 8)+(d + 8)(d + 8)\). Wait, \((d + 6)(d - 8)=d^2-8d + 6d-48=d^2-2d - 48\), \((d + 8)(d + 8)=d^2 + 16d+64\). Sum: \(d^2-2d - 48+d^2 + 16d + 64=2d^2+14d + 16\). Wait, but let's check the answer options. Option B is \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\), but wait, maybe I made a mistake in the initial factoring. Wait, wait, the original problem's numerator of the first fraction: \(d^2 + d - 30=(d + 6)(d - 5)\), denominator \(d^2 + 3d - 40=(d + 8)(d - 5)\), correct. Second fraction: numerator \(d^2 + 14d + 48=(d + 6)(d + 8)\), denominator \(d^2 - 2d - 48=(d - 8)(d + 6)\), correct. Then simplifying: first fraction \(\frac{d + 6}{d + 8}\), second \(\frac{d + 8}{d - 8}\). Then common denominator \((d + 8)(d - 8)\). Then \((d + 6)(d - 8)=d^2-8d + 6d-48=d^2-2d - 48\), \((d + 8)(d + 8)=d^2 + 16d + 64\). Sum: \(d^2-2d - 48 + d^2 + 16d + 64=2d^2 + 14d + 16\). Wait, but let's check the answer options again. Wait, option B is \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\), which matches our result. Wait, but earlier when I thought I made a mistake, maybe not. Wait, let's check the answer options again. Wait, the options:

A: \(\frac{2d^2 + 15d + 18}{2d^2 + d - 88}\)
B: \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\)
C: \(\frac{2d^2 + 15d + 18}{(d + 8)(d - 8)}\)
D: \(\frac{d^2 + 14d + 16}{(d + 8)(d - 8)}\)

Wait, our result is \(\frac{2d^2 + 14d + 16}{(d + 8)(d - 8)}\), which is option B. Wait, but let's re - check the addition of numerators:

\((d + 6)(d - 8)=d^2-8d + 6d-48=d^2-2d - 48\)
\((d + 8)(d + 8)=d^2 + 16d + 64\)
Adding them: \(d^2-2d - 48 + d^2 + 16d + 64=2d^2 + 14d + 16\). Yes, that's correct. So the answer should be option B.

Answer:

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