$\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0$

[ correction added, Thanks to @stef68 ]

$\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)$

5 0

3 years ago

How are the answers to problems 6,8,12, and 14 similar?how are they different?

Tasya [4]

Answer:

Step-by-step explanation:

8 0

3 years ago

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Midpoint of (-7,-14) and (-13,-6)

Sholpan [36]

Answer:

-10 HAPPY TO HELP ILY <3

Step-by-step explanation:

5 0

2 years ago

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A triangle has side lengths of 24 in., 32 in., and 40 in. Is the triangle acute, obtuse, or right? A) right. B) obtuse. C) acute

Feliz [49]

24, 32 , 40. Divide through by 8

3, 4, 5,

The 3 , 4, 5 reminds of the Pythagorean triple, which is a right angled triangle.

So a triangle with sides 24, 32, 40 is a right angled triangle.

Option A.

8 0

2 years ago

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