iohanetc

2021-08-16

To calculate: The product of $\left[x-\left(3-i\right)\right]\left[x-\left(3+i\right)\right]$

oppturf

Step 1
Polinomial identity:
1) $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
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(x) represents the give nexpression, then
$P\left(x\right)=\left[x-\left(3-i\right)\right]\left[3-\left(3+i\right)\right]$
$P\left(x\right)=\left(x-3+i\right)\left(x-3-i\right)$
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 expressionas,
$P\left(x\right)=\left(\left(x-3\right)+i\right)\left(\left(x-3\right)-i\right)$
Apply arithmetic rule.
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Here,

Hence,
$P\left(x\right)={\left(x-3\right)}^{2}-{i}^{2}$
Apply arithmetic rule:
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Here, NKS
Since ${\left(i\right)}^{2}=\left(-1\right)$, the expression becomes,
$P\left(x\right)={\left(x-3\right)}^{2}-{i}^{2}$
$={x}^{2}-6x+9+1$
$={x}^{2}-6x+10$
Hence, the product of $\left[x-\left(3-i\right)\right]\left[x-\left(3+i\right)\right]$ is ${x}^{2}-6x+10$

Jeffrey Jordon