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Factor completely. $162p^2-242$

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Integer addition and subtraction rules

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Q. Factor completely.

162p^2-242

Identify GCF: Identify the greatest common factor (GCF) of the terms $162p^2$ and $242$ . Both numbers are even, so $2$ is a factor. We can check for higher factors that are common to both numbers.
Divide by $2$ : Divide both $162$ and $242$ by $2$ to see if there is a larger common factor. $162 \div 2 = 81$ and $242 \div 2 = 121$ . Both $81$ and $121$ are perfect squares ( $9^2$ and $11^2$ , respectively), so the GCF is not just $2$ . We can try to factor out a larger number.
Check for $3$ : Since $81$ and $121$ are both multiples of $3$ , we can divide them by $3$ to see if $3$ is a common factor. $81 \div 3 = 27$ and $121$ is not divisible by $3$ . Therefore, $3$ is not a common factor for both terms. We should go back and check for the largest common factor that includes $2$ .
Factor out $2$ : We know that $81$ is $3^4$ and $121$ is $11^2$ . Since $3$ is not a factor of $121$ , we can conclude that the GCF is $2$ . Now we can factor out $2$ from both terms.
Factor inside parentheses: Factoring out the GCF of $2$ , we get $2(81p^2 - 121)$ . We can now look at each term inside the parentheses to see if they can be factored further.
Apply difference of squares: Inside the parentheses, we have a difference of squares since $81p^2$ is $(9p)^2$ and $121$ is $11^2$ . We can factor the difference of squares as $(a^2 - b^2) = (a + b)(a - b)$ .
Verify factoring: Apply the difference of squares factoring to get $2((9p + 11)(9p - 11))$ .
Verify factoring: Apply the difference of squares factoring to get $2((9p + 11)(9p - 11))$ .Check the factoring for any possible errors. $2 \times (9p + 11) \times (9p - 11)$ should expand back to the original expression $162p^2 - 242$ . Let's expand it to verify: $2 \times (81p^2 - 99p + 99p - 121)$ simplifies to $2 \times (81p^2 - 121)$ , which is $162p^2 - 242$ . The factoring is correct.

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