Question
Simplify:
$(2x + p - c)^2 - (2x - p + c)^2$
$(2x + p - c)^2 - (2x - p + c)^2$
✓
Answer
Given $(2x + p - c)^2 - (2x - p + c)^2$
By using identity $(x + y + z) = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$
we get $(2x + p - c)^2 - (2x - p + c)^2$
$=(2\text{x})^2+(\text{p})^2+(-\text{c})^2+2(2\text{x})(\text{p})+2(\text{p})(-\text{c})+2(-\text{c})(2\text{x})\\-\big[(2\text{x})^2+(-\text{p})^2+(\text{c})^2+2(2\text{x})(-\text{p})+2(-\text{p})(\text{c})+2(\text{c})(2\text{x})\big]$
$$$=4\text{x}^2+\text{p}^2+\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}\\-\big[4\text{x}^2+\text{p}^2+\text{c}^2-4\text{xp}-2\text{cp}+4\text{cx}\big]$
By cancelling the opposite terms,
we get$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2\\=4\text{x}^2+\not\text{p}^2+\not\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}-4\text{x}^2\\-\not\text{p}^2-\not\text{c}^2+4\text{xp}+2\text{cp}-4\text{cx}$
$=4\text{xp}+4\text{xp}-4\text{cx}-4\text{cx}$
$=8\text{xp}-8\text{cx}$
Talking $8x$ as common a factor we get,$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2=8\text{x}(\text{p}-\text{c})$
Hence the value of $\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2$ is $8\text{x}(\text{p}-\text{c}).$
By using identity $(x + y + z) = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$
we get $(2x + p - c)^2 - (2x - p + c)^2$
$=(2\text{x})^2+(\text{p})^2+(-\text{c})^2+2(2\text{x})(\text{p})+2(\text{p})(-\text{c})+2(-\text{c})(2\text{x})\\-\big[(2\text{x})^2+(-\text{p})^2+(\text{c})^2+2(2\text{x})(-\text{p})+2(-\text{p})(\text{c})+2(\text{c})(2\text{x})\big]$
$$$=4\text{x}^2+\text{p}^2+\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}\\-\big[4\text{x}^2+\text{p}^2+\text{c}^2-4\text{xp}-2\text{cp}+4\text{cx}\big]$
By cancelling the opposite terms,
we get$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2\\=4\text{x}^2+\not\text{p}^2+\not\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}-4\text{x}^2\\-\not\text{p}^2-\not\text{c}^2+4\text{xp}+2\text{cp}-4\text{cx}$
$=4\text{xp}+4\text{xp}-4\text{cx}-4\text{cx}$
$=8\text{xp}-8\text{cx}$
Talking $8x$ as common a factor we get,$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2=8\text{x}(\text{p}-\text{c})$
Hence the value of $\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2$ is $8\text{x}(\text{p}-\text{c}).$
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