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Question
factor the polynomial, if possible. drag the expressions into the box if they are part of the factored form of the polynomial. if the polynomial cannot be factored, drag prime. 36 - 100w²
Step1: Recognize difference of squares
The polynomial \(36 - 100w^2\) can be written as \(6^2-(10w)^2\), which is a difference of squares. The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\).
Step2: Apply the formula
Here, \(a = 6\) and \(b = 10w\). So, \(6^2-(10w)^2=(6 + 10w)(6 - 10w)\). We can also factor out the greatest common factor from each binomial. The GCF of 6 and 10 is 2, so \((6 + 10w)(6 - 10w)=2(3 + 5w)\times2(3 - 5w)=4(3 + 5w)(3 - 5w)\), but the basic factored form using the difference of squares is \((6 + 10w)(6 - 10w)\) or we can simplify further by factoring out 2 from each term: \(2(3 + 5w)\times2(3 - 5w)=4(3 + 5w)(3 - 5w)\), but the most straightforward factoring using the difference of squares is \((6 + 10w)(6 - 10w)\) or we can factor out 2 from each binomial to get \(4(3 + 5w)(3 - 5w)\). However, the initial factoring using the difference of squares formula gives \((6 + 10w)(6 - 10w)\) and we can also factor out 2 from each term: \(2(3 + 5w)\times2(3 - 5w)=4(3 + 5w)(3 - 5w)\). But the simplest factored form using the difference of squares is \((6 + 10w)(6 - 10w)\) or we can factor out 2 from each binomial: \(2(3 + 5w)\) and \(2(3 - 5w)\), so multiplying those gives \(4(3 + 5w)(3 - 5w)\). But the key is recognizing the difference of squares.
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The factored form of \(36 - 100w^2\) is \((6 + 10w)(6 - 10w)\) (or equivalently \(4(3 + 5w)(3 - 5w)\) after factoring out the GCF from each binomial).