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Galina-37 [17]
2 years ago
15

What are the real zeroes of x3 + 6 x2 – 9x - 54?

Mathematics
1 answer:
o-na [289]2 years ago
7 0

Answer:

Option B 3,-3,-6 is correct.

Step-by-step explanation:

We need to find real zeroes of x^3+6x^2-9x-54

Solving

x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)

Taking x^2 common from first 2 terms and -9 from last two terms we get

=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)\\

Taking (x+6) common

(x+6)(x^2-9)\\

x^2-9 can be solved using formula a^2-b^2 = (a+b)(a-b)

=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)

Putting it equal to zero,

(x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\,  x=3

So, Option B 3,-3,-6 is correct.

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Smaller angle = 41°

Larger angle = 139°

Step-by-step explanation:

(7x - 1) + (2x + 1) = 180

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