Question
Simplify the following products:$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
✓
Answer
In the given problem, we have to find product of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$ On rearranging we get$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)\\=\big[\text{x}^2+(\text{x}-2)\big]\big[\text{x}^2-(\text{x}-2)\big]$'
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)\\=\Big[+(\text{x}^2)^2-(\text{x}-2)^2\Big]$
$=\text{x}^4-\big(\text{x}^2-2\times2\times\text{x}+2^2\big)$
$=\text{x}^4-\big(\text{x}^2-4\text{x}+4\big)$
$=\text{x}^4-\text{x}^2+4\text{x}-4$
Hence the value of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$ is $\text{x}^4-\text{x}^2+4\text{x}-4.$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)\\=\Big[+(\text{x}^2)^2-(\text{x}-2)^2\Big]$
$=\text{x}^4-\big(\text{x}^2-2\times2\times\text{x}+2^2\big)$
$=\text{x}^4-\big(\text{x}^2-4\text{x}+4\big)$
$=\text{x}^4-\text{x}^2+4\text{x}-4$
Hence the value of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$ is $\text{x}^4-\text{x}^2+4\text{x}-4.$
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