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3 years ago

15

One zero of x^3-3x^2-6x+8=0 is -2. WHat are the other zeros of the function?

1 answer:

Irina18 [472]3 years ago

7 0

$x^3-3x^2-6x+8=x^3-x^2-2x^2+2x-8x+8=\]\[=x^2(x-1)-2x(x-1)-8(x-1)=(x-1)(x^2-2x-8)=\]\[=(x-1)(x^2+2x-4x-8)=(x-1)(x(x+2)-4(x+2))=\]\[=(x-1)(x+2)(x-4)$
Other zeros are 1 ana 4

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3 units

Step-by-step explanation:

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2 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

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Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

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Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

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$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

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3 years ago

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IceJOKER [234]

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