generals336

2021-08-11

To calculate: The product of $\left[x-\left(1+\sqrt{2}\right)\right]\left[x-\left(1-\sqrt{2}\right)\right]$.

wornoutwomanC

Step 1
Given:
Algebraic expression in multiplication form, $\left[x-\left(1+\sqrt{2}\right)\right]\left[x-\left(1-\sqrt{2}\right)\right]$
Formula Used:
Polynomial identity,
$\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$
$\left(a-b{\right)}^{2}={a}^{2}2ab+{b}^{2}$
Associative property, $\left(a+b\right)+c=a+\left(b+c\right)$
Step 2
If $P\left(x\right)$ represent the given expression, then
$P\left(x\right)=\left[x-\left(1+\sqrt{2}\right)\right]\left[x-\left(1-\sqrt{2}\right)\right]$
$P\left(x\right)=\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)$
Use associative property of algebraic expressions.
Associative property is written as.
$\left(a+b\right)+c=a+\left(b+c\right)$
This property modifies the expression as,
$P\left(x\right)=\left(\left(x-1\right)-\sqrt{2}\right)\left(\left(x-1\right)+\sqrt{2}\right)$
Apply arithmetic rule.
$\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$
Here,
Hence,
$P\left(x\right)={\left(x-1\right)}^{2}-\left({\sqrt{2}}^{2}$
Apply arithmetic rule:
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Here,
Hence,
$P\left(x\right)={x}^{2}-2x+1-2$
$={x}^{2}-2x-1$
Hence, the product of $\left[x-\left(1+\sqrt{2}\right)\right]\left[x-\left(1-\sqrt{2}\right)\right]$ is ${x}^{2}-2x-1.$

Jeffrey Jordon