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Yanka [14]
3 years ago
15

Given a polynomial and one of its factors bother main factors 9x^3 + 20x^2 - 68x - 16; (x+4)

Mathematics
1 answer:
icang [17]3 years ago
8 0

Answer:

9x+2 and x-2.

Step-by-step explanation:

Given the polynomial 9x^3 + 20x^2 - 68x - 16

If one of the factors is x+4, then to obtain the other factor, we divide the polynomial by the known factor.

\dfrac{9x^3 + 20x^2 - 68x - 16}{x+4} =9x^2-16x-4

Next, we factorize our result

9x^2-16x-4=9x^2-18x+2x-4\\=9x(x-2)+2(x-2)\\=(9x+2)(x-2)

Therefore, the other factors of the polynomial are 9x+2 and x-2.

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3 years ago
What is the value of x? 3x−8=8x+4/2 Enter your answer in the box
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For this case we must find the value of "x" of the following expression:

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Multiplying by 2 on both sides of the equation:

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Answer:

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It is given that,

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