Find the remainder when p(x)=8x^3-6x^2-4x+4 is divided by q(x)=x/3-1/4without performing the acual division.
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we use remainder theorem related to polynomials.
![p(x)=8x^3-6x^2-4x+4,\ \ \ q(x)=\frac{1}{3}(x-\frac{3}{4})\\\\\frac{p(x)}{q(x)}=3*\frac{p(x)}{x-\frac{3}{4}}=3*p(\frac{3}{4}),\ \ by\ remainder\ theorem.\\\\=3*[8(\frac{3}{4})^3-6(\frac{3}{4})^2-4*\frac{3}{4}+4]\\\\=3](https://tex.z-dn.net/?f=p%28x%29%3D8x%5E3-6x%5E2-4x%2B4%2C%5C+%5C+%5C+q%28x%29%3D%5Cfrac%7B1%7D%7B3%7D%28x-%5Cfrac%7B3%7D%7B4%7D%29%5C%5C%5C%5C%5Cfrac%7Bp%28x%29%7D%7Bq%28x%29%7D%3D3%2A%5Cfrac%7Bp%28x%29%7D%7Bx-%5Cfrac%7B3%7D%7B4%7D%7D%3D3%2Ap%28%5Cfrac%7B3%7D%7B4%7D%29%2C%5C+%5C+by%5C+remainder%5C+theorem.%5C%5C%5C%5C%3D3%2A%5B8%28%5Cfrac%7B3%7D%7B4%7D%29%5E3-6%28%5Cfrac%7B3%7D%7B4%7D%29%5E2-4%2A%5Cfrac%7B3%7D%7B4%7D%2B4%5D%5C%5C%5C%5C%3D3 "p(x)=8x^3-6x^2-4x+4,\ \ \ q(x)=\frac{1}{3}(x-\frac{3}{4})\\\\\frac{p(x)}{q(x)}=3*\frac{p(x)}{x-\frac{3}{4}}=3*p(\frac{3}{4}),\ \ by\ remainder\ theorem.\\\\=3*[8(\frac{3}{4})^3-6(\frac{3}{4})^2-4*\frac{3}{4}+4]\\\\=3")
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