Factoring Puzzle for Quadratic Trinomials | Math = Love - Free Printable

Let's solve the Factoring Puzzle step by step.

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🔍 Puzzle Instructions:

> Use the digits 0–9 to fill in the squares. Each digit can be used only once.

We have four quadratic trinomials, each with missing digits (in boxes), and we need to factor them correctly while using each digit from 0 to 9 exactly once.

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🧩 Problem Statements:

1. $ x^2 - x - \boxed{ } = (x + 2)(x - \boxed{ }) $
2. $ x^2 - 1\boxed{ }x + 1\boxed{ } = (x - 2)(x - \boxed{ }) $
3. $ x^2 + \boxed{ }x + 1\boxed{ } = (x + \boxed{ })(x + 2) $
4. $ x^2 - \boxed{ }x - 24 = (x - 6)(x + \boxed{ }) $

We'll go one at a time and use logic and algebra to determine the correct digits.

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## ✔ Step 1: Solve Equation (1)

$ x^2 - x - \boxed{ } = (x + 2)(x - \boxed{ }) $

Let’s expand the right-hand side:
$$
(x + 2)(x - a) = x^2 - ax + 2x - 2a = x^2 + (2 - a)x - 2a
$$

Compare to left-hand side: $ x^2 - x - b $

So:
- Coefficient of $ x $: $ 2 - a = -1 $ → $ a = 3 $
- Constant term: $ -2a = -b $ → $ -2(3) = -6 $ → $ b = 6 $

So:
- First blank: $ \boxed{6} $
- Second blank: $ \boxed{3} $

✔ So far:
- Used digits: 6, 3

Equation becomes:
$$
x^2 - x - 6 = (x + 2)(x - 3)
$$

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## ✔ Step 2: Solve Equation (2)

$ x^2 - 1\boxed{ }x + 1\boxed{ } = (x - 2)(x - \boxed{ }) $

Let’s call the missing digit in the linear term $ a $, and constant term $ b $. So:
$$
x^2 - 10a x + 10b = (x - 2)(x - c)
$$

Wait — actually, the format is:
$$
x^2 - 1\boxed{ }x + 1\boxed{ } = (x - 2)(x - d)
$$

So it's:
- Linear coefficient: $ 10a $? No — actually, the box is just one digit. So this means:
- The linear term is $ -1\boxed{}x $, meaning $ -10x $ or $ -15x $? Actually, no — the "1" is already written, so it's like $ -1d x $, where $ d $ is a single digit.

So let’s write it as:
$$
x^2 - (10 + d)x + (10 + e) = (x - 2)(x - f)
$$

Wait — better interpretation:
The expression is:
$$
x^2 - 1\boxed{}x + 1\boxed{} = (x - 2)(x - \boxed{})
$$

So:
- The linear term is $ -1d x $, where $ d $ is a digit → so coefficient is $ -(10 + d) $? No!

Wait — that would be if it were two-digit number.

But actually, the notation suggests:
- It's $ x^2 - 1\boxed{}x + 1\boxed{} $, meaning:
- The coefficient of $ x $ is a two-digit number starting with 1? Or is it $ -1 \times \text{digit} $?

No — look at the way it's written: `x² - 1□x + 1□`

That means:
- The coefficient of $ x $ is 10 + a, where $ a $ is the digit in the box.
- Similarly, the constant term is 10 + b, where $ b $ is the digit in the second box.

For example: if first box is 5, then coefficient is $ -15x $; if second box is 8, constant is $ +18 $.

So:
$$
x^2 - (10 + a)x + (10 + b) = (x - 2)(x - c)
$$

Now expand RHS:
$$
(x - 2)(x - c) = x^2 - (2 + c)x + 2c
$$

Set equal:
- $ -(10 + a) = -(2 + c) $ → $ 10 + a = 2 + c $ → $ c = 8 + a $
- $ 10 + b = 2c $

Substitute $ c = 8 + a $ into second equation:
$$
10 + b = 2(8 + a) = 16 + 2a \Rightarrow b = 6 + 2a
$$

Now $ a $ and $ b $ are digits from 0–9, and $ c $ must also be a digit (since it's a single box).

So:
- $ c = 8 + a \leq 9 $ → $ a \leq 1 $
- $ a \geq 0 $

Try $ a = 0 $:
- $ c = 8 $
- $ b = 6 + 0 = 6 $
- So equation: $ x^2 - 10x + 16 = (x - 2)(x - 8) $

Check:
$ (x - 2)(x - 8) = x^2 - 10x + 16 $ → YES!

Used digits: $ a=0 $, $ b=6 $, $ c=8 $

But wait — 6 is already used in equation (1). We can't reuse 6.

So invalid.

Try $ a = 1 $:
- $ c = 8 + 1 = 9 $
- $ b = 6 + 2(1) = 8 $
- So equation: $ x^2 - 11x + 18 = (x - 2)(x - 9) $

Check:
$ (x - 2)(x - 9) = x^2 - 11x + 18 $ → YES!

Digits used: $ a=1 $, $ b=8 $, $ c=9 $

So blanks:
- First box: 1 (in $ -1\boxed{1}x $)
- Second box: 8 (in $ +1\boxed{8} $)
- Third box: 9 (in $ x - \boxed{9} $)

Used digits: 1, 8, 9

Previously used: 3, 6

Now total used: 1, 3, 6, 8, 9

So far good.

✔ Equation (2):
$ x^2 - 11x + 18 = (x - 2)(x - 9) $

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## ✔ Step 3: Solve Equation (3)

$ x^2 + \boxed{ }x + 1\boxed{ } = (x + \boxed{ })(x + 2) $

Let’s denote:
- Let $ a $ be the digit in the linear term: $ +a x $
- Let $ b $ be the digit in the constant: $ +1b $ → so constant is $ 10 + b $
- Let $ c $ be the digit in $ (x + c) $

So:
$$
x^2 + a x + (10 + b) = (x + c)(x + 2)
$$

Expand RHS:
$$
x^2 + (c + 2)x + 2c
$$

Set equal:
- $ a = c + 2 $
- $ 10 + b = 2c $

From first: $ c = a - 2 $

Plug into second:
$$
10 + b = 2(a - 2) = 2a - 4 \Rightarrow b = 2a - 14
$$

Now $ a $, $ b $, $ c $ are digits from 0–9.

$ b = 2a - 14 \geq 0 $ → $ 2a \geq 14 $ → $ a \geq 7 $

Also $ b \leq 9 $ → $ 2a - 14 \leq 9 $ → $ 2a \leq 23 $ → $ a \leq 11.5 $ → $ a \leq 9 $

So possible $ a = 7, 8, 9 $

Try $ a = 7 $:
- $ b = 2(7) - 14 = 14 - 14 = 0 $
- $ c = a - 2 = 5 $
- So: $ x^2 + 7x + 10 = (x + 5)(x + 2) $

Check: $ (x+5)(x+2) = x^2 + 7x + 10 $ → YES!

Digits: $ a=7 $, $ b=0 $, $ c=5 $

Are these used?
Used so far: 1, 3, 6, 8, 9

New: 7, 0, 5 — all unused! Great.

So:
- First blank: $ \boxed{7} $
- Second blank: $ \boxed{0} $
- Third blank: $ \boxed{5} $

Now used digits: 0,1,3,5,6,7,8,9

Remaining digits: 2,4

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## ✔ Step 4: Solve Equation (4)

$ x^2 - \boxed{ }x - 24 = (x - 6)(x + \boxed{ }) $

Let’s denote:
- $ a $: digit in linear term → coefficient is $ -a $
- $ b $: digit in $ x + b $

So:
$$
x^2 - a x - 24 = (x - 6)(x + b)
$$

Expand RHS:
$$
x^2 + (b - 6)x - 6b
$$

Set equal:
- $ -a = b - 6 $ → $ a = 6 - b $
- $ -24 = -6b $ → $ b = 4 $

Then $ a = 6 - 4 = 2 $

So:
- $ b = 4 $
- $ a = 2 $

Now check digits:
- $ a = 2 $, $ b = 4 $

Available digits? Remaining: 2, 4 → perfect!

So:
- First blank: $ \boxed{2} $
- Second blank: $ \boxed{4} $

Check:
$ (x - 6)(x + 4) = x^2 + 4x - 6x - 24 = x^2 - 2x - 24 $ → YES!

Perfect.

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## ✔ Final Answers:

Now let’s fill in all the blanks.

Equation 1:

$ x^2 - x - \boxed{6} = (x + 2)(x - \boxed{3}) $

Equation 2:

$ x^2 - 1\boxed{1}x + 1\boxed{8} = (x - 2)(x - \boxed{9}) $

Equation 3:

$ x^2 + \boxed{7}x + 1\boxed{0} = (x + \boxed{5})(x + 2) $

Equation 4:

$ x^2 - \boxed{2}x - 24 = (x - 6)(x + \boxed{4}) $

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## ✔ Digits Used:
List of digits used:
- 0 → used in eq3
- 1 → used in eq2
- 2 → used in eq4
- 3 → used in eq1
- 4 → used in eq4
- 5 → used in eq3
- 6 → used in eq1
- 7 → used in eq3
- 8 → used in eq2
- 9 → used in eq2

✔ All digits 0–9 used exactly once.

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✔ Final Answer:

Here is the fully filled puzzle:

1. $ x^2 - x - \boxed{6} = (x + 2)(x - \boxed{3}) $
2. $ x^2 - 1\boxed{1}x + 1\boxed{8} = (x - 2)(x - \boxed{9}) $
3. $ x^2 + \boxed{7}x + 1\boxed{0} = (x + \boxed{5})(x + 2) $
4. $ x^2 - \boxed{2}x - 24 = (x - 6)(x + \boxed{4}) $

All equations are factored correctly, and digits 0–9 are used exactly once.

🎉 Puzzle Solved!