3 years ago

Explain how you see the x intercepts (also called roots) in the function y=(x-8)(x+10) without graphing it.

Mathematics

1 answer:

sveticcg [70]3 years ago

8 0

Answer:

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$\bold{\text{Answer:}\quad \dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}$

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$.\quad \dfrac{-5x}{8x+7}-\dfrac{6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}+\dfrac{-6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}\bigg(\dfrac{3x+1}{3x+1}\bigg)+\dfrac{-6x^3}{3x+1}\bigg(\dfrac{8x+7}{8x+7}\bigg)\\\\\\=\dfrac{-15x^2-5x}{(8x+7)(3x+1)}+\dfrac{-48x^4-42x^3}{(8x+7)(3x+1)}\\\\\\=\large\boxed{\dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}$

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Explain how you see the x intercepts (also called roots) in the function y=(x-8)(x+10) without graphing it.​

Explain how you see the x intercepts (also called roots) in the function y=(x-8)(x+10) without graphing it.