Factoring Trinomials worksheets - Free Printable

To solve the problem of factoring each trinomial, we will go through each expression step by step. Factoring trinomials involves expressing them as a product of two binomials (or other simpler factors). Here's how we approach each one:

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1) \( x^2 - 121 \)

This is a difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]

Here, \( a = x \) and \( b = 11 \):
\[ x^2 - 121 = (x - 11)(x + 11) \]

Answer:
\[ \boxed{(x - 11)(x + 11)} \]

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2) \( x^2 - 7x - 18 \)

We need to find two numbers that multiply to \(-18\) (the constant term) and add up to \(-7\) (the coefficient of \(x\)).

The numbers are \(-9\) and \(2\) because:
\[ -9 \cdot 2 = -18 \]
\[ -9 + 2 = -7 \]

So, we can factor as:
\[ x^2 - 7x - 18 = (x - 9)(x + 2) \]

Answer:
\[ \boxed{(x - 9)(x + 2)} \]

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3) \( x^2 + 5x - 36 \)

We need to find two numbers that multiply to \(-36\) and add up to \(5\).

The numbers are \(9\) and \(-4\) because:
\[ 9 \cdot (-4) = -36 \]
\[ 9 + (-4) = 5 \]

So, we can factor as:
\[ x^2 + 5x - 36 = (x + 9)(x - 4) \]

Answer:
\[ \boxed{(x + 9)(x - 4)} \]

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4) \( x^2 - 6x - 16 \)

We need to find two numbers that multiply to \(-16\) and add up to \(-6\).

The numbers are \(-8\) and \(2\) because:
\[ -8 \cdot 2 = -16 \]
\[ -8 + 2 = -6 \]

So, we can factor as:
\[ x^2 - 6x - 16 = (x - 8)(x + 2) \]

Answer:
\[ \boxed{(x - 8)(x + 2)} \]

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5) \( x^2 + 13x + 36 \)

We need to find two numbers that multiply to \(36\) and add up to \(13\).

The numbers are \(9\) and \(4\) because:
\[ 9 \cdot 4 = 36 \]
\[ 9 + 4 = 13 \]

So, we can factor as:
\[ x^2 + 13x + 36 = (x + 9)(x + 4) \]

Answer:
\[ \boxed{(x + 9)(x + 4)} \]

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6) \( -12x^2 - 26x - 12 \)

First, factor out the greatest common factor (\(-2\)):
\[ -12x^2 - 26x - 12 = -2(6x^2 + 13x + 6) \]

Now, factor \(6x^2 + 13x + 6\). We need two numbers that multiply to \(6 \cdot 6 = 36\) and add up to \(13\).

The numbers are \(9\) and \(4\) because:
\[ 9 \cdot 4 = 36 \]
\[ 9 + 4 = 13 \]

Rewrite the middle term using these numbers:
\[ 6x^2 + 13x + 6 = 6x^2 + 9x + 4x + 6 \]

Factor by grouping:
\[ = 3x(2x + 3) + 2(2x + 3) \]
\[ = (3x + 2)(2x + 3) \]

So, the full factorization is:
\[ -12x^2 - 26x - 12 = -2(3x + 2)(2x + 3) \]

Answer:
\[ \boxed{-2(3x + 2)(2x + 3)} \]

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7) \( x^2 - 6x + 5 \)

We need to find two numbers that multiply to \(5\) and add up to \(-6\).

The numbers are \(-5\) and \(-1\) because:
\[ -5 \cdot (-1) = 5 \]
\[ -5 + (-1) = -6 \]

So, we can factor as:
\[ x^2 - 6x + 5 = (x - 5)(x - 1) \]

Answer:
\[ \boxed{(x - 5)(x - 1)} \]

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8) \( x^2 + 10x + 25 \)

This is a perfect square trinomial:
\[ a^2 + 2ab + b^2 = (a + b)^2 \]

Here, \( a = x \) and \( b = 5 \):
\[ x^2 + 10x + 25 = (x + 5)^2 \]

Answer:
\[ \boxed{(x + 5)^2} \]

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9) \( x^2 - x - 20 \)

We need to find two numbers that multiply to \(-20\) and add up to \(-1\).

The numbers are \(-5\) and \(4\) because:
\[ -5 \cdot 4 = -20 \]
\[ -5 + 4 = -1 \]

So, we can factor as:
\[ x^2 - x - 20 = (x - 5)(x + 4) \]

Answer:
\[ \boxed{(x - 5)(x + 4)} \]

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10) \( x^2 - 6x + 5 \)

We need to find two numbers that multiply to \(5\) and add up to \(-6\).

The numbers are \(-5\) and \(-1\) because:
\[ -5 \cdot (-1) = 5 \]
\[ -5 + (-1) = -6 \]

So, we can factor as:
\[ x^2 - 6x + 5 = (x - 5)(x - 1) \]

Answer:
\[ \boxed{(x - 5)(x - 1)} \]

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11) \( x^2 + 4x - 45 \)

We need to find two numbers that multiply to \(-45\) and add up to \(4\).

The numbers are \(9\) and \(-5\) because:
\[ 9 \cdot (-5) = -45 \]
\[ 9 + (-5) = 4 \]

So, we can factor as:
\[ x^2 + 4x - 45 = (x + 9)(x - 5) \]

Answer:
\[ \boxed{(x + 9)(x - 5)} \]

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12) \( x^2 - 5x - 14 \)

We need to find two numbers that multiply to \(-14\) and add up to \(-5\).

The numbers are \(-7\) and \(2\) because:
\[ -7 \cdot 2 = -14 \]
\[ -7 + 2 = -5 \]

So, we can factor as:
\[ x^2 - 5x - 14 = (x - 7)(x + 2) \]

Answer:
\[ \boxed{(x - 7)(x + 2)} \]

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13) \( 2x^2 + 3x - 5 \)

We use the "ac method":
\[ a = 2, \, b = 3, \, c = -5 \]
\[ ac = 2 \cdot (-5) = -10 \]

We need two numbers that multiply to \(-10\) and add up to \(3\).

The numbers are \(5\) and \(-2\) because:
\[ 5 \cdot (-2) = -10 \]
\[ 5 + (-2) = 3 \]

Rewrite the middle term using these numbers:
\[ 2x^2 + 3x - 5 = 2x^2 + 5x - 2x - 5 \]

Factor by grouping:
\[ = x(2x + 5) - 1(2x + 5) \]
\[ = (x - 1)(2x + 5) \]

Answer:
\[ \boxed{(x - 1)(2x + 5)} \]

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14) \( x^2 + 8x + 15 \)

We need to find two numbers that multiply to \(15\) and add up to \(8\).

The numbers are \(5\) and \(3\) because:
\[ 5 \cdot 3 = 15 \]
\[ 5 + 3 = 8 \]

So, we can factor as:
\[ x^2 + 8x + 15 = (x + 5)(x + 3) \]

Answer:
\[ \boxed{(x + 5)(x + 3)} \]

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15) \( x^2 - 64 \)

This is a difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]

Here, \( a = x \) and \( b = 8 \):
\[ x^2 - 64 = (x - 8)(x + 8) \]

Answer:
\[ \boxed{(x - 8)(x + 8)} \]

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16) \( x^2 - 2x - 15 \)

We need to find two numbers that multiply to \(-15\) and add up to \(-2\).

The numbers are \(-5\) and \(3\) because:
\[ -5 \cdot 3 = -15 \]
\[ -5 + 3 = -2 \]

So, we can factor as:
\[ x^2 - 2x - 15 = (x - 5)(x + 3) \]

Answer:
\[ \boxed{(x - 5)(x + 3)} \]

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17) \( x^2 - 2x - 15 \)

We need to find two numbers that multiply to \(-15\) and add up to \(-2\).

The numbers are \(-5\) and \(3\) because:
\[ -5 \cdot 3 = -15 \]
\[ -5 + 3 = -2 \]

So, we can factor as:
\[ x^2 - 2x - 15 = (x - 5)(x + 3) \]

Answer:
\[ \boxed{(x - 5)(x + 3)} \]

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18) \( -10x^2 - 24x - 8 \)

First, factor out the greatest common factor (\(-2\)):
\[ -10x^2 - 24x - 8 = -2(5x^2 + 12x + 4) \]

Now, factor \(5x^2 + 12x + 4\). We need two numbers that multiply to \(5 \cdot 4 = 20\) and add up to \(12\).

The numbers are \(10\) and \(2\) because:
\[ 10 \cdot 2 = 20 \]
\[ 10 + 2 = 12 \]

Rewrite the middle term using these numbers:
\[ 5x^2 + 12x + 4 = 5x^2 + 10x + 2x + 4 \]

Factor by grouping:
\[ = 5x(x + 2) + 2(x + 2) \]
\[ = (5x + 2)(x + 2) \]

So, the full factorization is:
\[ -10x^2 - 24x - 8 = -2(5x + 2)(x + 2) \]

Answer:
\[ \boxed{-2(5x + 2)(x + 2)} \]

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19) \( x^2 - 9x + 18 \)

We need to find two numbers that multiply to \(18\) and add up to \(-9\).

The numbers are \(-6\) and \(-3\) because:
\[ -6 \cdot (-3) = 18 \]
\[ -6 + (-3) = -9 \]

So, we can factor as:
\[ x^2 - 9x + 18 = (x - 6)(x - 3) \]

Answer:
\[ \boxed{(x - 6)(x - 3)} \]

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20) \( x^2 - 4x - 5 \)

We need to find two numbers that multiply to \(-5\) and add up to \(-4\).

The numbers are \(-5\) and \(1\) because:
\[ -5 \cdot 1 = -5 \]
\[ -5 + 1 = -4 \]

So, we can factor as:
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]

Answer:
\[ \boxed{(x - 5)(x + 1)} \]

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21) \( x^2 - 11x + 18 \)

We need to find two numbers that multiply to \(18\) and add up to \(-11\).

The numbers are \(-9\) and \(-2\) because:
\[ -9 \cdot (-2) = 18 \]
\[ -9 + (-2) = -11 \]

So, we can factor as:
\[ x^2 - 11x + 18 = (x - 9)(x - 2) \]

Answer:
\[ \boxed{(x - 9)(x - 2)} \]

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22) \( x^2 - 9x + 14 \)

We need to find two numbers that multiply to \(14\) and add up to \(-9\).

The numbers are \(-7\) and \(-2\) because:
\[ -7 \cdot (-2) = 14 \]
\[ -7 + (-2) = -9 \]

So, we can factor as:
\[ x^2 - 9x + 14 = (x - 7)(x - 2) \]

Answer:
\[ \boxed{(x - 7)(x - 2)} \]

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Final Answer:

\[
\boxed{
\begin{aligned}
1) & \ (x - 11)(x + 11) \\
2) & \ (x - 9)(x + 2) \\
3) & \ (x + 9)(x - 4) \\
4) & \ (x - 8)(x + 2) \\
5) & \ (x + 9)(x + 4) \\
6) & \ -2(3x + 2)(2x + 3) \\
7) & \ (x - 5)(x - 1) \\
8) & \ (x + 5)^2 \\
9) & \ (x - 5)(x + 4) \\
10) & \ (x - 5)(x - 1) \\
11) & \ (x + 9)(x - 5) \\
12) & \ (x - 7)(x + 2) \\
13) & \ (x - 1)(2x + 5) \\
14) & \ (x + 5)(x + 3) \\
15) & \ (x - 8)(x + 8) \\
16) & \ (x - 5)(x + 3) \\
17) & \ (x - 5)(x + 3) \\
18) & \ -2(5x + 2)(x + 2) \\
19) & \ (x - 6)(x - 3) \\
20) & \ (x - 5)(x + 1) \\
21) & \ (x - 9)(x - 2) \\
22) & \ (x - 7)(x - 2)
\end{aligned}
}
\]