Factoring Trinomials Worksheets - Math Monks - Free Printable

Let’s solve each trinomial step by step. We’re factoring trinomials of the form ax² + bx + c, where a ≠ 1.

We’ll use the “AC method” (also called grouping):

Steps:

1. Multiply a × c.
2. Find two numbers that multiply to a×c and add to b.
3. Rewrite the middle term using those two numbers.
4. Factor by grouping.
5. Check your answer by multiplying back.

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Problem 1: 5x² + 26x + 24

- a = 5, b = 26, c = 24 → a×c = 5×24 = 120
- Need two numbers that multiply to 120 and add to 26 → 20 and 6
- Rewrite: 5x² + 20x + 6x + 24
- Group: (5x² + 20x) + (6x + 24)
- Factor: 5x(x + 4) + 6(x + 4)
- Final: (5x + 6)(x + 4)

✔ Check: (5x+6)(x+4) = 5x² + 20x + 6x + 24 = 5x² + 26x + 24 ✔️

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Problem 2: 2x² + 7x - 30

- a=2, b=7, c=-30 → a×c = -60
- Need two numbers that multiply to -60 and add to 7 → 12 and -5
- Rewrite: 2x² + 12x - 5x - 30
- Group: (2x² + 12x) + (-5x - 30)
- Factor: 2x(x + 6) -5(x + 6)
- Final: (2x - 5)(x + 6)

✔ Check: (2x-5)(x+6) = 2x² + 12x -5x -30 = 2x² +7x -30 ✔️

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Problem 3: 3x² - 8x + 4

- a=3, b=-8, c=4 → a×c = 12
- Need two numbers that multiply to 12 and add to -8 → -6 and -2
- Rewrite: 3x² -6x -2x + 4
- Group: (3x² -6x) + (-2x + 4)
- Factor: 3x(x - 2) -2(x - 2)
- Final: (3x - 2)(x - 2)

✔ Check: (3x-2)(x-2) = 3x² -6x -2x +4 = 3x² -8x +4 ✔️

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Problem 4: 2x² + 5x + 2

- a=2, b=5, c=2 → a×c = 4
- Numbers that multiply to 4 and add to 5 → 4 and 1
- Rewrite: 2x² + 4x + x + 2
- Group: (2x² + 4x) + (x + 2)
- Factor: 2x(x + 2) + 1(x + 2)
- Final: (2x + 1)(x + 2)

✔ Check: (2x+1)(x+2) = 2x² +4x +x +2 = 2x² +5x +2 ✔️

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Problem 5: 6x² - 10x + 50

Wait — let’s check if we can factor out a GCF first.

All coefficients are even? 6, -10, 50 → yes! GCF = 2

Factor out 2:
→ 2(3x² - 5x + 25)

Now try to factor 3x² -5x +25:

a=3, b=-5, c=25 → a×c = 75

Need two numbers that multiply to 75 and add to -5 → possible pairs: (-1,-75), (-3,-25), (-5,-15)... none add to -5.

Discriminant: b² - 4ac = (-5)² - 4(3)(25) = 25 - 300 = -275 → negative → no real factors.

So this trinomial is prime after factoring out GCF.

Final Answer: 2(3x² - 5x + 25) — cannot be factored further over integers.

But wait — maybe I made a mistake? Let me double-check original problem: 6x² -10x +50.

Yes, GCF is 2, and inside is prime. So final factored form is just with GCF.

Actually — let me check discriminant again for 3x² -5x +25:

Δ = 25 - 300 = -275 → definitely not factorable over reals as integer binomials.

So answer is: 2(3x² - 5x + 25)

BUT — sometimes problems expect you to leave it unfactored if not possible. However, since instruction says “factor completely”, and we factored out GCF, that’s complete.

Wait — perhaps typo in problem? Because 6x² -10x +50 seems odd. But assuming it's correct...

Alternatively, maybe they meant 6x² -10x -50? But we go with what’s given.

So: 2(3x² - 5x + 25) — but actually, let me see if 3x² -5x +25 can be factored... no, because no integer pair multiplies to 75 and adds to -5.

So final answer for #5: 2(3x² - 5x + 25)

But hold on — let me check online or think again... Actually, maybe I should proceed and come back.

Wait — perhaps I miscalculated a×c? For 3x² -5x +25, a×c=75, yes. Pairs: 1&75, 3&25, 5&15 — all positive, so to get sum -5, need both negative: -1 & -75 → sum -76; -3 & -25 → -28; -5 & -15 → -20 — none give -5. So indeed, not factorable.

So #5: 2(3x² - 5x + 25)

But let’s move on and come back if needed.

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Problem 6: 3x² - 2x - 5

- a=3, b=-2, c=-5 → a×c = -15
- Need two numbers that multiply to -15 and add to -2 → -5 and 3
- Rewrite: 3x² -5x + 3x -5
- Group: (3x² -5x) + (3x -5)
- Factor: x(3x -5) +1(3x -5)
- Final: (x + 1)(3x - 5)

✔ Check: (x+1)(3x-5) = 3x² -5x +3x -5 = 3x² -2x -5 ✔️

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Problem 7: 5x² -14x +8

- a=5, b=-14, c=8 → a×c=40
- Need two numbers that multiply to 40 and add to -14 → -10 and -4
- Rewrite: 5x² -10x -4x +8
- Group: (5x² -10x) + (-4x +8)
- Factor: 5x(x - 2) -4(x - 2)
- Final: (5x - 4)(x - 2)

✔ Check: (5x-4)(x-2) = 5x² -10x -4x +8 = 5x² -14x +8 ✔️

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Problem 8: 9x² +66x +21

First, check GCF: 9, 66, 21 → divisible by 3.

Factor out 3: 3(3x² + 22x + 7)

Now factor 3x² +22x +7:

a=3, b=22, c=7 → a×c=21

Numbers that multiply to 21 and add to 22 → 21 and 1

Rewrite: 3x² +21x + x +7

Group: (3x² +21x) + (x +7)

Factor: 3x(x +7) +1(x +7)

→ (3x +1)(x +7)

So overall: 3(3x + 1)(x + 7)

✔ Check: 3*(3x+1)(x+7) = 3*[3x² +21x +x +7] = 3*[3x² +22x +7] = 9x² +66x +21 ✔️

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Problem 9: -6x² +5x +6

It’s easier to factor out -1 first to make leading coefficient positive.

→ -1(6x² -5x -6)

Now factor 6x² -5x -6:

a=6, b=-5, c=-6 → a×c = -36

Need two numbers that multiply to -36 and add to -5 → -9 and 4

Rewrite: 6x² -9x +4x -6

Group: (6x² -9x) + (4x -6)

Factor: 3x(2x -3) +2(2x -3)

→ (3x +2)(2x -3)

So overall: -1(3x + 2)(2x - 3) or written as -(3x + 2)(2x - 3)

You could also write as ( -3x -2 )(2x -3) but better to keep the negative sign outside.

Standard way: -(3x + 2)(2x - 3)

✔ Check: -(3x+2)(2x-3) = -[6x² -9x +4x -6] = -[6x² -5x -6] = -6x² +5x +6 ✔️

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Problem 10: 7x² +53x +28

a=7, b=53, c=28 → a×c=196

Need two numbers that multiply to 196 and add to 53.

What pairs multiply to 196?

1×196 → sum 197
2×98 → 100
4×49 → 53 ← YES!

So 4 and 49.

Rewrite: 7x² +49x +4x +28

Group: (7x² +49x) + (4x +28)

Factor: 7x(x +7) +4(x +7)

→ (7x + 4)(x + 7)

✔ Check: (7x+4)(x+7) = 7x² +49x +4x +28 = 7x² +53x +28 ✔️

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Problem 11: -6x² + x + 12

Again, factor out -1 to make leading coefficient positive:

→ -1(6x² - x - 12)

Now factor 6x² - x -12:

a=6, b=-1, c=-12 → a×c = -72

Need two numbers that multiply to -72 and add to -1 → -9 and 8

Rewrite: 6x² -9x +8x -12

Group: (6x² -9x) + (8x -12)

Factor: 3x(2x -3) +4(2x -3)

→ (3x +4)(2x -3)

So overall: -1(3x + 4)(2x - 3) or -(3x + 4)(2x - 3)

✔ Check: -(3x+4)(2x-3) = -[6x² -9x +8x -12] = -[6x² -x -12] = -6x² +x +12 ✔️

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Problem 12: 4x² -4x -63

a=4, b=-4, c=-63 → a×c = -252

Need two numbers that multiply to -252 and add to -4.

Let’s list factor pairs of 252:

1×252
2×126
3×84
4×63
6×42
7×36
9×28
12×21
14×18 → difference is 4? 18 -14 =4 → so if we take -18 and +14 → sum = -4, product = -252 ✔️

So numbers: -18 and 14

Rewrite: 4x² -18x +14x -63

Group: (4x² -18x) + (14x -63)

Factor: 2x(2x -9) +7(2x -9)

→ (2x + 7)(2x - 9)

✔ Check: (2x+7)(2x-9) = 4x² -18x +14x -63 = 4x² -4x -63 ✔️

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Now back to Problem 5: 6x² -10x +50

We said GCF is 2 → 2(3x² -5x +25)

And 3x² -5x +25 has discriminant = 25 - 300 = -275 < 0 → not factorable over reals.

But perhaps the problem expects us to leave it as is? Or maybe there’s a typo? Since other problems factor nicely, maybe it’s supposed to be 6x² -10x -50? Let’s check that quickly.

If it were 6x² -10x -50:

GCF=2 → 2(3x² -5x -25)

Then a×c=3*(-25)= -75

Need two numbers multiply to -75, add to -5 → -15 and 5? -15+5=-10 no. -10 and 7.5? Not integer.

Discriminant: 25 + 300 = 325 → not perfect square.

Or maybe 6x² -11x +50? No.

Perhaps it’s 6x² -10x +4? Then GCF=2 → 2(3x² -5x +2) → which factors to 2(3x-2)(x-1)? Let’s not guess.

Given the problem as written, we must factor completely — so factoring out GCF is the only thing we can do.

So for #5: 2(3x² - 5x + 25)

But let me confirm once more: is 3x² -5x +25 factorable? Suppose (3x+a)(x+b) = 3x² + (3b+a)x + ab

Set ab=25, 3b+a=-5

Possible a,b: ±1,±5,±25

Try a=-5, b=-5: ab=25, 3*(-5)+(-5)= -15-5=-20 ≠ -5

a=-25, b=-1: ab=25, 3*(-1)+(-25)= -3-25=-28

a=5, b=5: 3*5+5=20

a=25, b=1: 3*1+25=28

a=-1, b=-25: 3*(-25)+(-1)= -75-1=-76

No integer solutions. So yes, not factorable.

Thus, final answer for #5 is 2(3x² - 5x + 25)

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Now compiling all answers:

1) (5x + 6)(x + 4)
2) (2x - 5)(x + 6)
3) (3x - 2)(x - 2)
4) (2x + 1)(x + 2)
5) 2(3x² - 5x + 25)
6) (x + 1)(3x - 5)
7) (5x - 4)(x - 2)
8) 3(3x + 1)(x + 7)
9) -(3x + 2)(2x - 3)
10) (7x + 4)(x + 7)
11) -(3x + 4)(2x - 3)
12) (2x + 7)(2x - 9)

Note: For #9 and #11, some might prefer to distribute the negative into one binomial, but standard practice is to leave the negative sign outside unless specified.

For example, #9: -(3x+2)(2x-3) is fine. Alternatively, ( -3x -2 )(2x -3) or (3x+2)(-2x+3), but the first form is clearest.

Similarly for #11.

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Final Answer:

1) (5x + 6)(x + 4)
2) (2x - 5)(x + 6)
3) (3x - 2)(x - 2)
4) (2x + 1)(x + 2)
5) 2(3x² - 5x + 25)
6) (x + 1)(3x - 5)
7) (5x - 4)(x - 2)
8) 3(3x + 1)(x + 7)
9) -(3x + 2)(2x - 3)
10) (7x + 4)(x + 7)
11) -(3x + 4)(2x - 3)
12) (2x + 7)(2x - 9)