Multiplying Binomials Using The FOIL Method: - Free Printable

To solve the given binomials using the FOIL method, we will follow these steps:

FOIL Method Recap:

- First: Multiply the first terms in each binomial.
- Outside: Multiply the outer terms in the product.
- Inside: Multiply the inner terms in the product.
- Last: Multiply the last terms in each binomial.

After applying FOIL, combine like terms to simplify the expression.

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Problem 1: \((x + 2)(2x + 5)\)

1. First: \(x \cdot 2x = 2x^2\)
2. Outside: \(x \cdot 5 = 5x\)
3. Inside: \(2 \cdot 2x = 4x\)
4. Last: \(2 \cdot 5 = 10\)

Combine all terms:
\[
2x^2 + 5x + 4x + 10 = 2x^2 + 9x + 10
\]

Answer: \(\boxed{2x^2 + 9x + 10}\)

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Problem 2: \((3x - 3)(x + 3)\)

1. First: \(3x \cdot x = 3x^2\)
2. Outside: \(3x \cdot 3 = 9x\)
3. Inside: \(-3 \cdot x = -3x\)
4. Last: \(-3 \cdot 3 = -9\)

Combine all terms:
\[
3x^2 + 9x - 3x - 9 = 3x^2 + 6x - 9
\]

Answer: \(\boxed{3x^2 + 6x - 9}\)

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Problem 3: \((4a - 5)(a + 3)\)

1. First: \(4a \cdot a = 4a^2\)
2. Outside: \(4a \cdot 3 = 12a\)
3. Inside: \(-5 \cdot a = -5a\)
4. Last: \(-5 \cdot 3 = -15\)

Combine all terms:
\[
4a^2 + 12a - 5a - 15 = 4a^2 + 7a - 15
\]

Answer: \(\boxed{4a^2 + 7a - 15}\)

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Problem 4: \((2a + 1)(a + 3)\)

1. First: \(2a \cdot a = 2a^2\)
2. Outside: \(2a \cdot 3 = 6a\)
3. Inside: \(1 \cdot a = a\)
4. Last: \(1 \cdot 3 = 3\)

Combine all terms:
\[
2a^2 + 6a + a + 3 = 2a^2 + 7a + 3
\]

Answer: \(\boxed{2a^2 + 7a + 3}\)

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Problem 5: \((2x + 7)(x - 8)\)

1. First: \(2x \cdot x = 2x^2\)
2. Outside: \(2x \cdot (-8) = -16x\)
3. Inside: \(7 \cdot x = 7x\)
4. Last: \(7 \cdot (-8) = -56\)

Combine all terms:
\[
2x^2 - 16x + 7x - 56 = 2x^2 - 9x - 56
\]

Answer: \(\boxed{2x^2 - 9x - 56}\)

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Problem 6: \((x + 2)^2\)

This is a special case of squaring a binomial. We can write it as:
\[
(x + 2)(x + 2)
\]

1. First: \(x \cdot x = x^2\)
2. Outside: \(x \cdot 2 = 2x\)
3. Inside: \(2 \cdot x = 2x\)
4. Last: \(2 \cdot 2 = 4\)

Combine all terms:
\[
x^2 + 2x + 2x + 4 = x^2 + 4x + 4
\]

Answer: \(\boxed{x^2 + 4x + 4}\)

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Problem 7: \((3x + 1)(2x + 6)\)

1. First: \(3x \cdot 2x = 6x^2\)
2. Outside: \(3x \cdot 6 = 18x\)
3. Inside: \(1 \cdot 2x = 2x\)
4. Last: \(1 \cdot 6 = 6\)

Combine all terms:
\[
6x^2 + 18x + 2x + 6 = 6x^2 + 20x + 6
\]

Answer: \(\boxed{6x^2 + 20x + 6}\)

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Problem 8: \((a - 5)^2\)

This is another special case of squaring a binomial. We can write it as:
\[
(a - 5)(a - 5)
\]

1. First: \(a \cdot a = a^2\)
2. Outside: \(a \cdot (-5) = -5a\)
3. Inside: \(-5 \cdot a = -5a\)
4. Last: \(-5 \cdot (-5) = 25\)

Combine all terms:
\[
a^2 - 5a - 5a + 25 = a^2 - 10a + 25
\]

Answer: \(\boxed{a^2 - 10a + 25}\)

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Final Answers:

1. \((x + 2)(2x + 5) = \boxed{2x^2 + 9x + 10}\)
2. \((3x - 3)(x + 3) = \boxed{3x^2 + 6x - 9}\)
3. \((4a - 5)(a + 3) = \boxed{4a^2 + 7a - 15}\)
4. \((2a + 1)(a + 3) = \boxed{2a^2 + 7a + 3}\)
5. \((2x + 7)(x - 8) = \boxed{2x^2 - 9x - 56}\)
6. \((x + 2)^2 = \boxed{x^2 + 4x + 4}\)
7. \((3x + 1)(2x + 6) = \boxed{6x^2 + 20x + 6}\)
8. \((a - 5)^2 = \boxed{a^2 - 10a + 25}\)