5.1: Multiplying and Special Products | Algebra 1 Honors

How it works

Multiply every term in one factor by every term in the other, then combine like terms. Squaring a sum creates a middle term; it is not the same as squaring each term separately. The product of conjugate binomials cancels the middle terms.

(a+b)^2=a^2+2ab+b^2,\qquad(a-b)(a+b)=a^2-b^2

Worked examples

Example 1

Expand $\displaystyle (x+3)(x-2)$ .

Distribute: $\displaystyle x^2-2x+3x-6=x^2+x-6$ .

Example 2

Expand $\displaystyle (2x-1)^2$ .

Use the square formula: $\displaystyle (2x)^2-2(2x)(1)+1=4x^2-4x+1$ .

Problems

Problem 5.1.1

Expand $\displaystyle (x+4)^2$ .

Show worked solution

$\displaystyle x^2+2(x)(4)+16=x^2+8x+16$ .

Problem 5.1.2

Expand $\displaystyle (3x+2)(x-5)$ .

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$\displaystyle 3x^2-15x+2x-10=3x^2-13x-10$ .

Problem 5.1.3

Simplify $\displaystyle (x-7)(x+7)$ .

Show worked solution

The conjugate factors give a difference of squares: $\displaystyle x^2-49$ .

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 5.1.1

Problem 5.1.2

Problem 5.1.3

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