Work through these rational expression simplification problems to strengthen your algebra skills

Let’s solve each rational expression one by one. We’ll factor the numerator and denominator, then cancel any common factors.

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1. $\frac{n^2 + 7n + 6}{n^2 - n - 42}$

Factor numerator:
$n^2 + 7n + 6 = (n+1)(n+6)$

Factor denominator:
$n^2 - n - 42 = (n-7)(n+6)$

Cancel $(n+6)$:
→ $\frac{n+1}{n-7}$

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2. $\frac{h^2 + 13h + 42}{h + 7}$

Factor numerator:
$h^2 + 13h + 42 = (h+6)(h+7)$

Denominator is $h+7$

Cancel $(h+7)$:
→ $h+6$

*(Note: This assumes $h \ne -7$, but since we’re simplifying, we just write the simplified form.)*

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3. $\frac{s^2 - 4s - 45}{s^2 + 2s - 15}$

Factor numerator:
$s^2 - 4s - 45 = (s-9)(s+5)$

Factor denominator:
$s^2 + 2s - 15 = (s+5)(s-3)$

Cancel $(s+5)$:
→ $\frac{s-9}{s-3}$

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4. $\frac{g + 9}{g^2 + 16g + 63}$

Factor denominator:
$g^2 + 16g + 63 = (g+7)(g+9)$

Numerator is $g+9$

Cancel $(g+9)$:
→ $\frac{1}{g+7}$

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5. $\frac{c^2 - 8c + 15}{c - 3}$

Factor numerator:
$c^2 - 8c + 15 = (c-3)(c-5)$

Denominator is $c-3$

Cancel $(c-3)$:
→ $c-5$

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6. $\frac{8h^2 + 15h - 2}{4h^2 + 11h + 6}$

Factor numerator:
We need two numbers that multiply to $8 \cdot (-2) = -16$ and add to 15 → 16 and -1? No. Try factoring by grouping or trial.

Try:
$(8h - 1)(h + 2) = 8h^2 + 16h - h - 2 = 8h^2 + 15h - 2$ ✔

Factor denominator:
$4h^2 + 11h + 6$ → look for two numbers multiplying to $4\cdot6=24$, adding to 11 → 8 and 3.

So:
$4h^2 + 8h + 3h + 6 = 4h(h+2) + 3(h+2) = (4h+3)(h+2)$

Now expression becomes:
$\frac{(8h - 1)(h + 2)}{(4h + 3)(h + 2)}$

Cancel $(h+2)$:
→ $\frac{8h - 1}{4h + 3}$

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7. $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$

Check if numerator factors:
Discriminant: $(-4)^2 - 4(3)(-6) = 16 + 72 = 88$ → not a perfect square → doesn’t factor nicely over integers.

Denominator: $5r^2 - 16r + 3$
Discriminant: $(-16)^2 - 4(5)(3) = 256 - 60 = 196 = 14^2$ → does factor!

Find roots:
$r = \frac{16 \pm 14}{10} = \frac{30}{10}=3$ or $\frac{2}{10}=0.2$

So factors: $(5r - 1)(r - 3)$? Let's check:
$(5r - 1)(r - 3) = 5r^2 -15r -r +3 = 5r^2 -16r +3$ ✔

But numerator $3r^2 -4r -6$ — let’s try to factor it anyway.

Looking for two numbers multiplying to $3*(-6)=-18$, adding to -4 → possible pairs: (-6,3), (-9,2), etc. None work.

Wait — maybe I made a mistake? Let me double-check discriminant of numerator:

$3r^2 -4r -6$: discriminant = $16 + 72 = 88$ → sqrt(88)=2√22 → irrational → so no integer factors.

That means this fraction cannot be simplified further using integer factoring.

BUT wait — let me check if there’s a common factor between numerator and denominator.

Numerator: $3r^2 -4r -6$

Denominator: $(5r - 1)(r - 3)$

Does numerator share either factor?

Test r=3 in numerator: $3(9) -4(3) -6 = 27 -12 -6 = 9 ≠ 0$ → not divisible by (r-3)

Test r=1/5: $3(1/25) -4(1/5) -6 = 3/25 - 20/25 - 150/25 = -167/25 ≠ 0$

So no common factors → already simplified.

Wait — did I misread the problem? Let me recheck original:

Original: $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$

Yes — and denominator factors to $(5r-1)(r-3)$, numerator doesn't factor with integers and shares no common linear factors → so answer is itself? But that seems odd for a “simplify” worksheet.

Wait — perhaps I made an error in numerator factoring attempt.

Alternative approach: Maybe typo? Or maybe I should leave as is?

Actually, let me try again on numerator: $3r^2 -4r -6$

Try AC method: A=3, C=-6 → AC=-18

Need two numbers multiply to -18, add to -4 → possibilities:

-6 and 3 → sum -3 ✘
-9 and 2 → sum -7 ✘
-18 and 1 → sum -17 ✘
No pair adds to -4 → so indeed, doesn’t factor over integers.

Therefore, this expression is already in simplest form.

But let’s hold that thought — maybe I missed something.

Wait — what if the numerator was meant to be $3r^2 - 4r - 4$? Then it would factor. But as written, it’s -6.

Alternatively, perhaps the problem expects us to say “already simplified”.

But let’s move on and come back.

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8. $\frac{10k^2 - 29k + 21}{2k^2 - 7k + 6}$

Factor numerator:
$10k^2 -29k +21$

AC method: 10*21=210, find two numbers multiply to 210, add to -29 → negative both.

Factors of 210: 14 and 15 → 14+15=29 → so -14 and -15.

Split middle term:
$10k^2 -15k -14k +21 = 5k(2k - 3) -7(2k - 3) = (5k - 7)(2k - 3)$ ✔

Denominator: $2k^2 -7k +6$

AC: 2*6=12, numbers multiply to 12, add to -7 → -3 and -4

Split: $2k^2 -4k -3k +6 = 2k(k-2) -3(k-2) = (2k-3)(k-2)$ ✔

So expression:
$\frac{(5k - 7)(2k - 3)}{(2k - 3)(k - 2)}$

Cancel $(2k - 3)$:
→ $\frac{5k - 7}{k - 2}$

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Now go back to #7: $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$

Since numerator doesn’t factor over integers and shares no common factors with denominator, it is already simplified.

But let me verify once more — perhaps I can use quadratic formula to see if they share a root.

Numerator roots: $r = \frac{4 \pm \sqrt{16 + 72}}{6} = \frac{4 \pm \sqrt{88}}{6} = \frac{4 \pm 2\sqrt{22}}{6} = \frac{2 \pm \sqrt{22}}{3}$

Denominator roots: from earlier, r=3 and r=1/5

None match → so no common factors.

Thus, final answer for #7 is the same as original.

But that feels unusual for a simplification worksheet — perhaps there’s a typo in the problem? However, based on what’s given, we must proceed.

Alternatively, maybe I miscalculated denominator factoring?

Denominator: $5r^2 -16r +3$

We had: (5r -1)(r -3) = 5r² -15r -r +3 = 5r² -16r +3 → correct.

Numerator: 3r² -4r -6 — let’s plug in r=3: 27 -12 -6=90; r=1/5: 3/25 -4/5 -6 = 3/25 -20/25 -150/25 = -167/25≠0 → confirmed.

So yes, cannot simplify.

Perhaps the problem intended $3r^2 - 4r - 4$? That would factor as (3r+2)(r-2). But as written, it’s -6.

I’ll proceed with current values.

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Final Answers:

1. $\frac{n+1}{n-7}$
2. $h+6$
3. $\frac{s-9}{s-3}$
4. $\frac{1}{g+7}$
5. $c-5$
6. $\frac{8h - 1}{4h + 3}$
7. $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$ *(already simplified)*
8. $\frac{5k - 7}{k - 2}$

Wait — for #7, maybe the problem has a typo, but since we must solve as given, we leave it.

However, let me check online or think differently — sometimes "simplify" includes stating domain restrictions, but the instruction is to simplify the expression, meaning reduce algebraically.

Given that, and since no cancellation is possible, we leave it.

But let me try one last thing — perhaps factor numerator by grouping differently?

3r² -4r -6 — group as (3r² -6) -4r = 3(r² -2) -4r — no help.

Or (3r² -4r) -6 = r(3r-4) -6 — still no.

So yes, irreducible over integers.

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Final Answer:

1. $\frac{n+1}{n-7}$
2. $h+6$
3. $\frac{s-9}{s-3}$
4. $\frac{1}{g+7}$
5. $c-5$
6. $\frac{8h - 1}{4h + 3}$
7. $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$
8. $\frac{5k - 7}{k - 2}$

But for #7, since it's listed among others that simplify, perhaps I made a mistake.

Wait — let me recalculate discriminant of numerator: 3r² -4r -6

a=3, b=-4, c=-6

discriminant = b² -4ac = 16 - 4(3)(-6) = 16 + 72 = 88 — correct.

sqrt(88) = 2√22 — not rational — so no rational roots — so no linear factors with integer coefficients — so cannot cancel with denominator which has rational roots.

Thus, it is already simplified.

Perhaps the answer key expects to leave it as is.

I'll go with that.

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Final Answer:
1. $\frac{n+1}{n-7}$
2. $h+6$
3. $\frac{s-9}{s-3}$
4. $\frac{1}{g+7}$
5. $c-5$
6. $\frac{8h - 1}{4h + 3}$
7. $\frac{3r^2 - 4r - 6}{5r^2 - 16r + 3}$
8. $\frac{5k - 7}{k - 2}$