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Popular Calculus >

expand \sqrt[3]{7x^2+7x+5}*(5x-3)

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Solution

expand 37x2+7x+5​⋅(5x−3)

Solution

5x37x2+7x+5​−337x2+7x+5​
Solution steps
37x2+7x+5​(5x−3)
Apply the distributive law: a(b−c)=ab−ac37x2+7x+5​(5x−3)=37x2+7x+5​⋅5x−37x2+7x+5​⋅3=37x2+7x+5​⋅5x−37x2+7x+5​⋅3
=5x37x2+7x+5​−337x2+7x+5​

Popular Examples

integral of 1/(3cos(5x-pi/4))∫3cos(5x−4π​)1​dxintegral of arctan(y)∫arctan(y)dycos^2(x)sin(x)((dy)/(dx))+(cos^3(x))y=1cos2(x)sin(x)(dxdy​)+(cos3(x))y=1integral of ((sin(x))/x)∫(xsin(x)​)dx(dy)/(dt)=-\sqrt[3]{y}dtdy​=−3y​

Frequently Asked Questions (FAQ)

  • What is expand \sqrt[3]{7x^2+7x+5}*(5x-3) ?

    The solution to expand \sqrt[3]{7x^2+7x+5}*(5x-3) is 5x\sqrt[3]{7x^2+7x+5}-3\sqrt[3]{7x^2+7x+5}
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