Question

1. factor $36 - 4x^2$ completely.

Explanation:

Step1: Factor out the GCF

First, find the greatest common factor (GCF) of \(36\) and \(4x^2\). The GCF of \(36\) and \(4\) is \(4\)? Wait, no, wait: \(36 = 4\times9\) and \(4x^2=4\times x^2\)? Wait, no, actually, \(36 = 2^2\times9\) and \(4x^2 = 2^2\times x^2\). Wait, the GCF of \(36\) and \(4x^2\) is \(4\)? Wait, \(36\div4 = 9\), \(4x^2\div4 = x^2\). Wait, no, \(36 - 4x^2 = 4(9 - x^2)\)? Wait, no, \(36\) is \(6^2\), \(4x^2\) is \((2x)^2\). Wait, maybe first factor out the common factor. Let's see: \(36 - 4x^2 = 4(9) - 4(x^2) = 4(9 - x^2)\). Then, \(9 - x^2\) is a difference of squares, since \(9 = 3^2\) and \(x^2 = x^2\). The formula for difference of squares is \(a^2 - b^2=(a - b)(a + b)\). So \(9 - x^2 = 3^2 - x^2=(3 - x)(3 + x)\). Therefore, putting it all together: \(36 - 4x^2 = 4(9 - x^2)=4(3 - x)(3 + x)\). Wait, but also, we can factor out the 4 first, or notice that \(36 - 4x^2=(6)^2-(2x)^2\), which is a difference of squares directly. Let's try that. The difference of squares formula is \(a^2 - b^2=(a - b)(a + b)\). So here, \(a = 6\), \(b = 2x\). So \(36 - 4x^2 = 6^2 - (2x)^2=(6 - 2x)(6 + 2x)\). Then, we can factor out the common factor from each binomial. From \(6 - 2x\), factor out 2: \(2(3 - x)\). From \(6 + 2x\), factor out 2: \(2(3 + x)\). Then, multiply the 2s together: \(2\times2 = 4\). So \((6 - 2x)(6 + 2x)=2(3 - x)\times2(3 + x)=4(3 - x)(3 + x)\). So both methods give the same result. Let's do it step by step.

Step1: Identify the difference of squares

First, rewrite the expression as a difference of squares. \(36 - 4x^2 = 6^2 - (2x)^2\).

Step2: Apply the difference of squares formula

Using \(a^2 - b^2=(a - b)(a + b)\), where \(a = 6\) and \(b = 2x\), we get:
\(6^2 - (2x)^2=(6 - 2x)(6 + 2x)\)

Step3: Factor out the common factor from each binomial

From \(6 - 2x\), factor out 2: \(6 - 2x = 2(3 - x)\)
From \(6 + 2x\), factor out 2: \(6 + 2x = 2(3 + x)\)

Step4: Multiply the factored forms

Now, multiply \(2(3 - x)\) and \(2(3 + x)\):
\(2(3 - x)\times2(3 + x)=4(3 - x)(3 + x)\)

Answer:

\(4(3 - x)(3 + x)\) (or also written as \(4(3 + x)(3 - x)\) since multiplication is commutative)