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Question

Using division algorithm, find quotient and remainder dividing $$f(x)$$ by $$g(x)$$.
$$f(x)=9x^4-4x^2+4,g(x)=3x^2+x-1$$

Solution
Verified by Toppr

Given $$f(x)=9x^4-4x^2+4$$
and $$g(x)=3x^2+x-1$$
Dividing $$f(x)$$ by $$g(x)$$
$$\quad \quad \quad \quad \quad \quad \quad \quad \quad 3x^{ 2 }-x\\ \left. 3x^{ 2 }+x-1 \right) \overline { 9x^{ 4 }-4x^{ 2 }+4 } \\ \quad \quad \quad \quad \quad \,\, 9x^{ 4 }-3x^{ 3 } -3x^{ 2 }\\ \quad \quad \quad \quad \_ -\_ \_ +\_ \_ \_ \_ \_ +\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad -3x^{ 3 }-x^{ 2 }+4\\ \quad \quad \quad \quad \quad \quad \quad \quad -3x^{ 3 }-x^{ 2 }\quad +x\\ \quad \quad \quad \quad \quad \quad \quad \quad \_ +\_ \_ \_ +\_ \_ \_ \_ -\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -x+4$$
Thus quotient $$q(x)=3x^2-x$$
Remainder $$f(x)=-x+4$$
Here:$$f(x).q(x)+f(x)$$
$$=(3x^2+x-1)(3x^2-x)+(-x)+4$$
$$=9x^4+3x^3-3x^2-3x^3-x^2+x-x+4$$
$$=9x^4-4x^2+4$$
$$=f(x)$$
Thus, division algorithm is verified.

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