Factoring Polynomials Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve the factoring by grouping problems step-by-step, starting with the main example and then moving on to the additional exercises.

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🔹 Definition Fill-in:

> grouping - a method of factoring when a polynomial has four terms

✔ So, the completed sentence is:
> *Grouping – a method of factoring when a polynomial has four terms.*

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## 🎯 MAIN EXAMPLE:
Factor:
$$
12x^2 + 2x - 30x - 5
$$

Wait — let’s simplify first. Combine like terms:

$$
12x^2 + (2x - 30x) - 5 = 12x^2 - 28x - 5
$$

But this isn't how we're supposed to do it. The original expression was written as:
$$
12x^2 + 2x - 30x - 5
$$
So we keep it as four terms and factor by grouping.

Let’s go through the steps:

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✔ Step 1: Factor out the GCF from the first two terms

First two terms: $12x^2 + 2x$
GCF = $2x$
$$
2x(6x + 1)
$$

✔ Step 2: Factor out the GCF from the second two terms

Next two terms: $-30x - 5$
GCF = $-5$
$$
-5(6x + 1)
$$

✔ Step 3: Put the common remainder in one parenthesis; the two GCFs in another

Now we have:
$$
2x(6x + 1) - 5(6x + 1)
$$

Both groups have a common binomial factor: $(6x + 1)$

So factor that out:
$$
(6x + 1)(2x - 5)
$$

✔ Final Answer:
$$
\boxed{(6x + 1)(2x - 5)}
$$

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Now let's go through the other examples using the same method.

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## 🔹 Problem 1: $2x^2 + x + 8x + 4$

Combine like terms: $2x^2 + 9x + 4$, but again, we’ll group as is.

Step 1: First two terms: $2x^2 + x$

GCF = $x$ → $x(2x + 1)$

Step 2: Next two terms: $8x + 4$

GCF = $4$ → $4(2x + 1)$

Step 3: Now we have:

$$
x(2x + 1) + 4(2x + 1)
= (2x + 1)(x + 4)
$$

✔ Final Answer:
$$
\boxed{(2x + 1)(x + 4)}
$$

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## 🔹 Problem 2: $2x^2 - 14x + 4x - 28$

Simplify: $2x^2 - 10x - 28$ — but again, use grouping.

Step 1: $2x^2 - 14x$ → GCF = $2x$ → $2x(x - 7)$

Step 2: $4x - 28$ → GCF = $4$ → $4(x - 7)$

Step 3:

$$
2x(x - 7) + 4(x - 7) = (x - 7)(2x + 4)
$$

Now check if we can factor further: $2x + 4 = 2(x + 2)$

So:
$$
(x - 7)(2x + 4) = 2(x - 7)(x + 2)
$$

✔ Final Answer:
$$
\boxed{2(x - 7)(x + 2)}
$$

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## 🔹 Problem 3: $15x^2 + 5x + 3x + 1$

Step 1: $15x^2 + 5x$ → GCF = $5x$ → $5x(3x + 1)$

Step 2: $3x + 1$ → GCF = $1$ → $1(3x + 1)$

Step 3:

$$
5x(3x + 1) + 1(3x + 1) = (3x + 1)(5x + 1)
$$

✔ Final Answer:
$$
\boxed{(3x + 1)(5x + 1)}
$$

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## 🔹 Problem 4: $2x^3 - 16x^2 + 10x - 80$

This is a four-term polynomial, so we can group.

Step 1: First two terms: $2x^3 - 16x^2$ → GCF = $2x^2$ → $2x^2(x - 8)$

Step 2: Next two terms: $10x - 80$ → GCF = $10$ → $10(x - 8)$

Step 3:

$$
2x^2(x - 8) + 10(x - 8) = (x - 8)(2x^2 + 10)
$$

Now factor out GCF from $2x^2 + 10$: $2(x^2 + 5)$

So:
$$
(x - 8)(2x^2 + 10) = 2(x - 8)(x^2 + 5)
$$

✔ Final Answer:
$$
\boxed{2(x - 8)(x^2 + 5)}
$$

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## 🔹 Problem 5: $6x^3 - 16x^2 + 21x - 56$

Group into two pairs:

Step 1: $6x^3 - 16x^2$ → GCF = $2x^2$ → $2x^2(3x - 8)$

Step 2: $21x - 56$ → GCF = $7$ → $7(3x - 8)$

Step 3:

$$
2x^2(3x - 8) + 7(3x - 8) = (3x - 8)(2x^2 + 7)
$$

✔ Final Answer:
$$
\boxed{(3x - 8)(2x^2 + 7)}
$$

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## 🔹 Problem 6: $x^3 - 9x^2 + 6x - 54$

Group:

Step 1: $x^3 - 9x^2$ → GCF = $x^2$ → $x^2(x - 9)$

Step 2: $6x - 54$ → GCF = $6$ → $6(x - 9)$

Step 3:

$$
x^2(x - 9) + 6(x - 9) = (x - 9)(x^2 + 6)
$$

✔ Final Answer:
$$
\boxed{(x - 9)(x^2 + 6)}
$$

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## ✔ Summary of All Answers:

| Problem | Factored Form |
|--------|----------------|
| 1. $2x^2 + x + 8x + 4$ | $(2x + 1)(x + 4)$ |
| 2. $2x^2 - 14x + 4x - 28$ | $2(x - 7)(x + 2)$ |
| 3. $15x^2 + 5x + 3x + 1$ | $(3x + 1)(5x + 1)$ |
| 4. $2x^3 - 16x^2 + 10x - 80$ | $2(x - 8)(x^2 + 5)$ |
| 5. $6x^3 - 16x^2 + 21x - 56$ | $(3x - 8)(2x^2 + 7)$ |
| 6. $x^3 - 9x^2 + 6x - 54$ | $(x - 9)(x^2 + 6)$ |

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✔ Key Tips for Factoring by Grouping:

1. Always look at four-term polynomials.
2. Group into two pairs.
3. Factor out the GCF from each pair.
4. Look for a common binomial factor.
5. If none exists, try regrouping or check for errors.

Let me know if you'd like these explained visually or with diagrams!