$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

2 years ago

What is the equation of the line passing through the points (4,8) and (-1,-12)

d1i1m1o1n [39]

Answer:

y=4x-8 hope this helps:)

6 0

2 years ago

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