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#mathematics #polynomials #precalculus

Calculate \(299\times 301\).

You can brute force the answer to this problem by using a calculator, but we have a sweeter way. We can apply the difference of two squares identity.

At first we may think about using the long multiplication method, but it wastes time and is, of course, boring. Notice that \(299=300-1\) and \(301=300+1\), so

\(299\times 301=(300-1)(300+1)=300^2-1^2=89999\).

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**Difference Of Squares | Brilliant Math & Science Wiki**

es identity: \(a^2-b^2=(a+b)(a-b)\). Here are the examples to learn the usage of the identity. Rewrite \(5^2-2^2\) as a product. We have \[5^2-2^2 = (5-2) \times (5+2) = 3\times 7. \ _\square\] <span>Calculate \(299\times 301\). You can brute force the answer to this problem by using a calculator, but we have a sweeter way. We can apply the difference of two squares identity. At first we may think about using the long multiplication method, but it wastes time and is, of course, boring. Notice that \(299=300-1\) and \(301=300+1\), so \[\begin{align*}299\times 301&=(300-1)(300+1)\\&=300^2-1^2\\&=89999. \ _\square \end{align*}\] Show that any odd number can be written as the difference of two squares. Let the odd number be \( n = 2b + 1 \), where \(b\) is a non-negative integer. Then we have \[ n = 2b+1 = [ (b+

es identity: \(a^2-b^2=(a+b)(a-b)\). Here are the examples to learn the usage of the identity. Rewrite \(5^2-2^2\) as a product. We have \[5^2-2^2 = (5-2) \times (5+2) = 3\times 7. \ _\square\] <span>Calculate \(299\times 301\). You can brute force the answer to this problem by using a calculator, but we have a sweeter way. We can apply the difference of two squares identity. At first we may think about using the long multiplication method, but it wastes time and is, of course, boring. Notice that \(299=300-1\) and \(301=300+1\), so \[\begin{align*}299\times 301&=(300-1)(300+1)\\&=300^2-1^2\\&=89999. \ _\square \end{align*}\] Show that any odd number can be written as the difference of two squares. Let the odd number be \( n = 2b + 1 \), where \(b\) is a non-negative integer. Then we have \[ n = 2b+1 = [ (b+

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