Question
Simplify the following products:
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
✓
Answer
In the given problem, we have to find product of
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
On rearranging we get $\Big(\big[2\text{x}^4-4\text{x}^2\big]+1\Big)\Big(\big[2\text{x}^4-4\text{x}^2\big]-1\Big)$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y)}=\text{x}^2-\text{y}^2$
$\big(2\text{x}^4-4\text{x}^2+1\big)\big(2\text{x}^4-4\text{x}^2-1\big)\\=\big[2\text{x}^4-4\text{x}^2\big]^2-1^2$
$=\big[4\text{x}^8+16\text{x}^4-2\times2\text{x}^4\times4\text{x}^2-1\big]$
$=4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
Hence the value of $(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$ is $4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
On rearranging we get $\Big(\big[2\text{x}^4-4\text{x}^2\big]+1\Big)\Big(\big[2\text{x}^4-4\text{x}^2\big]-1\Big)$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y)}=\text{x}^2-\text{y}^2$
$\big(2\text{x}^4-4\text{x}^2+1\big)\big(2\text{x}^4-4\text{x}^2-1\big)\\=\big[2\text{x}^4-4\text{x}^2\big]^2-1^2$
$=\big[4\text{x}^8+16\text{x}^4-2\times2\text{x}^4\times4\text{x}^2-1\big]$
$=4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
Hence the value of $(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$ is $4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
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