Question

factor completely: $x^2 - 9$
simplify: $(x^2 + 3x)(x - 2)$

Explanation:

Part 1: Factor completely \( x^2 - 9 \)

Step 1: Identify the formula

Notice that \( x^2 - 9 \) is a difference of squares, which follows the formula \( a^2 - b^2=(a + b)(a - b) \). Here, \( a = x \) and \( b = 3 \) (since \( 9 = 3^2 \)).

Step 2: Apply the formula

Substitute \( a=x \) and \( b = 3 \) into the difference - of - squares formula: \( x^2-9=x^2 - 3^2=(x + 3)(x - 3) \)

Part 2: Simplify \( (x^2+3x)(x - 2) \)

Step 1: Factor out the GCF from the first binomial

First, factor out the greatest common factor (GCF) from \( x^2+3x \). The GCF of \( x^2 \) and \( 3x \) is \( x \), so \( x^2 + 3x=x(x + 3) \).

Step 2: Multiply the two binomials

Now, we have \( x(x + 3)(x - 2) \). We can also expand the original product \( (x^2+3x)(x - 2) \) using the distributive property (FOIL method for binomials, extended for a binomial and a trinomial).
Multiply \( x^2 \) by \( x-2 \): \( x^2\times x=x^3 \), \( x^2\times(-2)=-2x^2 \)
Multiply \( 3x \) by \( x - 2 \): \( 3x\times x = 3x^2 \), \( 3x\times(-2)=-6x \)
Then combine like terms: \( x^3-2x^2+3x^2-6x=x^3+x^2-6x \) (or we can leave it in factored form as \( x(x + 3)(x - 2) \))

Answer:

s:

• Factored form of \( x^2-9 \): \( (x + 3)(x - 3) \)
• Simplified form of \( (x^2+3x)(x - 2) \): \( x^3+x^2-6x \) (or \( x(x + 3)(x - 2) \))