vaderlord203
Answered

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find all the real and complex zeros of f(x)=x^3-7x^2+4x-28

Sagot :

mirai123

Answer:

For [tex]x \in\mathbb{C}[/tex]

[tex]x = -2i[/tex]

[tex]x = 2i[/tex]

[tex]x = 7[/tex]

Step-by-step explanation:

[tex]f(x)=x^3-7x^2+4x-28[/tex]

For [tex]f(x) = 0[/tex]

[tex]x^3-7x^2+4x-28 = 0[/tex]

We can easily factor the expression [tex]x^3-7x^2+4x-28[/tex] by grouping.

[tex]x^3-7x^2+4x-28 = (x^3-7x^2)+(4x-28) = x^2(x-7)+4(x-7) = \boxed{(x^2+4)(x-7) }[/tex]

Now we have

[tex](x^2+4)(x-7) = 0[/tex]

Therefore,

[tex]x^2+4 = 0 \text{ or } x-7 = 0[/tex]

So,

[tex]x^2 = -4 \implies x = \pm\sqrt{-4}[/tex]

Once [tex]\sqrt{-1} = i[/tex]

[tex]x = \pm\sqrt{-4} = \pm\sqrt{4}\sqrt{-1} = \pm 2i[/tex]

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