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$$\frac{-52y^6 - 96y^5 + 24y^4 - 68y^3}{4y^3}$$ enter the correct answe…

Question

$$\frac{-52y^6 - 96y^5 + 24y^4 - 68y^3}{4y^3}$$ enter the correct answer.

Explanation:

Step1: Divide each term by \(4y^3\)

For the first term: \(\frac{-52y^6}{4y^3} = -13y^{6 - 3} = -13y^3\)
For the second term: \(\frac{-96y^5}{4y^3} = -24y^{5 - 3} = -24y^2\)
For the third term: \(\frac{24y^4}{4y^3} = 6y^{4 - 3} = 6y\)
For the fourth term: \(\frac{-68y^3}{4y^3} = -17\)

Step2: Combine the results

Putting all the terms together, we get \(-13y^3 - 24y^2 + 6y - 17\)

Answer:

\(-13y^3 - 24y^2 + 6y - 17\)