Adding Mixed Numbers Worksheets - Free Printable

Let’s solve each of the 4 factoring problems one by one. We’ll look for two numbers that multiply to give the constant term (the last number) and add up to the middle coefficient (the number in front of x).

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Problem 1: x² + 8x + 7

We need two numbers that:
- Multiply to 7
- Add to 8

Factors of 7: 1 and 7 → 1 × 7 = 7, and 1 + 7 = 8 ✔

So, this factors as:
(x + 1)(x + 7)

Looking at the suspects list — Mrs. Peacock has (x + 7)(x + 1) — same thing! So she’s a suspect.

But wait — we’re told to eliminate suspects whose expressions match the answer. Since this matches Mrs. Peacock, we eliminate her.

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Problem 2: x² - 11x + 10

Need two numbers that:
- Multiply to 10
- Add to -11

Since the product is positive and sum is negative, both numbers must be negative.

Factors of 10: (-1) and (-10) → (-1) × (-10) = 10, (-1) + (-10) = -11 ✔

So, this factors as:
(x - 1)(x - 10)

Check weapons: Candlestick is (x - 10)(x - 1) — same thing! Eliminate Candlestick.

Also check suspects: Mr. Green has (x + 1)(x - 1) — not the same. Prof. Plum is “not factorable” — no. Coln. Mustard? Not listed here. Miss Scarlet? No. Mrs. White? (x - 2)(x - 3) — no.

Wait — actually, let’s double-check: Is there any suspect with (x - 1)(x - 10)? Looking again… none of the suspects have that exact pair. But wait — maybe I misread.

Actually, looking back: The only weapon matching is Candlestick. Suspects don’t seem to match this one directly. Let’s keep going.

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Problem 3: x² + x - 90

Need two numbers that:
- Multiply to -90
- Add to +1

One positive, one negative. Which pair adds to 1?

Try: 10 and -9 → 10 × (-9) = -90, 10 + (-9) = 1 ✔

So, this factors as:
(x + 10)(x - 9)

Check weapons: Knife is (x - 10)(x + 9) — close but signs are swapped. That would be x² - x - 90 — not the same.

Wait — what about Lead Pipe? It says (x + 6)(x - 9) — that’s x² - 3x - 54 — nope.

Hmm — maybe I made a mistake? Let me recalculate:

(x + 10)(x - 9) = x² -9x +10x -90 = x² + x - 90 ✔ Correct.

Is there any weapon or suspect with this? Doesn’t look like it from the image description. Maybe it doesn’t match anyone — so we don’t eliminate anyone for this one? Or perhaps I missed something.

Wait — let’s hold on and do problem 4 first.

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Problem 4: x² + 4x - 12

Need two numbers that:
- Multiply to -12
- Add to +4

Possible pairs: 6 and -2 → 6 × (-2) = -12, 6 + (-2) = 4 ✔

So, this factors as:
(x + 6)(x - 2)

Check weapons: Lead Pipe is (x + 6)(x - 9) — no, that’s different.

Wait — Lead Pipe is (x+6)(x-9), which is x² -3x -54 — not this.

What about suspects? Mrs. White has (x - 2)(x - 3) — that’s x² -5x +6 — no.

Miss Scarlet has (x - 11)(x - 1) — x² -12x +11 — no.

Coln. Mustard? Not shown in text.

Wait — maybe I need to re-express (x + 6)(x - 2). Is that listed anywhere?

Looking back at original image description:

Weapons:
- Candlestick: (x–10)(x–1)
- Knife: (x–10)(x+9)
- Lead Pipe: (x+6)(x–9)

None match (x+6)(x–2).

Suspects:
Mr. Green: (x+1)(x–1)
Prof. Plum: not factorable
Coln. Mustard: (x+8)(x+1)
Mrs. Peacock: (x+7)(x+1)
Miss Scarlet: (x–11)(x–1)
Mrs. White: (x–2)(x–3)

Still no match for (x+6)(x–2).

Wait — did I make an error in problem 3?

Problem 3 was x² + x - 90 → (x+10)(x-9)

Is that listed? Weapons: Knife is (x–10)(x+9) — which is x² -x -90 — opposite sign on middle term.

So not matching.

Perhaps some problems don’t correspond to any suspect/weapon — meaning those people/items are NOT eliminated.

The goal is to find who DIDN’T get eliminated — because they’re the culprit.

Let’s tabulate what we’ve found:

From Problem 1: x² + 8x + 7 = (x+1)(x+7) → matches Mrs. Peacock → eliminate her.

From Problem 2: x² -11x +10 = (x-1)(x-10) → matches Candlestick → eliminate Candlestick.

From Problem 3: x² + x -90 = (x+10)(x-9) → does NOT match any listed suspect or weapon → so NO elimination.

From Problem 4: x² +4x -12 = (x+6)(x-2) → does NOT match any listed suspect or weapon → NO elimination.

Wait — but maybe I missed something. Let me check if any suspect has (x+10)(x-9) or (x+6)(x-2). From the given list, no.

But perhaps Coln. Mustard? He has (x+8)(x+1) — not matching.

Maybe the key is that only some problems lead to eliminations.

Alternatively — perhaps I made a mistake in assuming all four problems must eliminate someone. The instruction says: “Factor each problem and find your answer on the grid. Eliminate that suspect, location or weapon...”

So if the factored form isn't on the grid, you don’t eliminate anyone for that problem.

Therefore, after solving all four:

Eliminated:
- Mrs. Peacock (from prob 1)
- Candlestick (from prob 2)

Not eliminated:
- All other suspects: Mr. Green, Prof. Plum, Coln. Mustard, Miss Scarlet, Mrs. White
- Other weapons: Knife, Lead Pipe
- Locations? Wait — the image mentions “location” too, but in the text provided, only suspects and weapons are listed. Maybe locations are implied elsewhere, but since not given, we assume only suspects and weapons matter.

But the question is: “who the dastardly villain is that messed up their math assignment!”

So we need to find which suspect remains uneliminated — but multiple remain.

Wait — perhaps I missed a match.

Let me go back to Problem 3: x² + x - 90 = (x+10)(x-9)

Is there a weapon called “Knife” with (x–10)(x+9)? That’s different — it’s (x-10)(x+9) = x² -x -90, whereas we have x² +x -90.

So not the same.

But what if the order doesn’t matter? (x+10)(x-9) vs (x-9)(x+10) — same thing. Still not matching Knife’s (x-10)(x+9).

Unless... is there a typo? Or perhaps I should consider absolute values? No.

Another thought: Maybe for Problem 4, (x+6)(x-2) — is that similar to Lead Pipe’s (x+6)(x-9)? Close but not same.

Perhaps the intended match is elsewhere.

Wait — let's look at Mrs. White: (x-2)(x-3) — that’s for x² -5x +6.

Not matching.

Miss Scarlet: (x-11)(x-1) = x² -12x +11.

No.

Mr. Green: (x+1)(x-1) = x² -1.

No.

Coln. Mustard: (x+8)(x+1) = x² +9x +8.

No.

Prof. Plum: not factorable — so he can't be matched to any factored expression — so he should NOT be eliminated? Because his expression isn't factorable, while all our problems are factorable.

Ah! That’s a clue.

The instruction says: “Someone didn’t factor right! The horror! Your job is to find out who, with what and where.”

And “Eliminate that suspect, location or weapon to find out who the dastardly villain is”

Also, Prof. Plum is labeled “not factorable” — but all four problems ARE factorable. So if Prof. Plum claims his expression is not factorable, but in reality, the problems are factorable, then HE might be the one who messed up — because he said it wasn’t factorable when it actually is.

That makes sense!

Let me verify:

All four quadratics we solved are factorable:

1. x² +8x+7 = (x+1)(x+7) — factorable
2. x² -11x+10 = (x-1)(x-10) — factorable
3. x² +x-90 = (x+10)(x-9) — factorable
4. x² +4x-12 = (x+6)(x-2) — factorable

Prof. Plum is marked as “not factorable” — implying he thinks his expression can’t be factored, but since all these can be factored, he is incorrect — so he is the villain who messed up.

Moreover, in the elimination process, we eliminate those whose expressions MATCH the answers. Prof. Plum’s expression is “not factorable”, which doesn’t match any of the factored forms, so he wouldn’t be eliminated by matching — but the twist is that he claimed it wasn’t factorable when it actually is (in the context of the problems given).

The story says: “Someone didn’t factor right!” — meaning someone made a mistake in factoring. Prof. Plum said “not factorable”, but if the actual problems are factorable, then he is wrong — so he is the culprit.

Additionally, in many such puzzles, the “not factorable” label is a red herring or indicates the person who failed to factor correctly.

So, putting it all together:

- We eliminate suspects/weapons whose factored forms match the solutions.
- Mrs. Peacock and Candlestick are eliminated.
- Prof. Plum is not eliminated by matching, but he is the one who incorrectly stated his expression was not factorable — when in fact, the problems show that factoring is possible.
- Therefore, he is the villain who messed up.

Final Answer: Prof. Plum