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Mathematics

2 answers:

aleksandrvk [35]3 years ago

8 0

Answer:

(x - 5)( x + 2i)(x - 2i)

Step-by-step explanation:

There might be a real root which is a factor of 20

Try 4:-

f(4) = 4^3 - 5*16 + 16 - 20 = -20 so its not 4

f(5) = 125 - 125 + 20 -20 = 0 so 5 is a root.

If 5 is a root then (x - 5) is a factor of the function:-

If we divide f(x) by x-5 we get

x^2 + 4

factor this- we get (x + 2i)(x - 2i)

so the factors are (x - 5)( x + 2i)(x - 2i)

uranmaximum [27]3 years ago

5 0

$f(x) = x^3 - 5x^2 + 4x - 20 \\\\f(x)=x^2(x-5)+4(x-5)\\\\f(x)=(x^2+4)(x-5)\\\\f(x)=(x-2i)(x+2i)(x-5)$

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grigory [225]

Answer:

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Step-by-step explanation:

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Rewriting

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