If $f(x)=x^2+bx+c$ where b,c are Real number and $f(x)$ is a factor of both $x^4+6x^2+25$ & $3x^4+4x^2+28x+5$ then find the value of $f(x)$.
I manage to get the answer by dividing the function by $f(x)$ and substituting the remainder as zero. I managed to get the following
$(x^2-2x+5)(x^2+2x+5)=x^4+6x^2+25$
$(x^2-2x+5)(3x^2+6x+1)=3x^4+4x^2+28x+5$
but i am not satisfied as it is a tedious process.
Is there any short-cut method
On
$$x^4+6x^2+25=x^4+10x^2+25-4x^2=(x^2+5)^2-(2x)^2=(x^2+2x+5)(x^2-2x+5).$$ Now, you can check that just $x^2-2x+5$ is a factor of the second polynomial.
If you want factoring of the second without this hint, I am ready to show, but it's not so easy.
On
Using Euclidean algorithm:
\begin{align} \frac{3x^4+4x^2+28x+5}{x^4+6x^2+25} &= 3+\frac{\color{red}{-14x^2+28x-70}}{x^4+6x^2+25} \\ \frac{x^4+6x^2+25}{\color{red}{-14(x^2-2x+5)}} &= -\frac{x^2+2x+5}{14} \end{align}
RHS is a pure polynomial (i.e. without rational function as remainder), therefore GCD is $$\color{red}{x^2-2x+5}$$
$$ \left( x^{4} + 6 x^{2} + 25 \right) $$
$$ \left( 3 x^{4} + 4 x^{2} + 28 x + 5 \right) $$
$$ \left( x^{4} + 6 x^{2} + 25 \right) = \left( 3 x^{4} + 4 x^{2} + 28 x + 5 \right) \cdot \color{magenta}{ \left( \frac{ 1}{3 } \right) } + \left( \frac{ 14 x^{2} - 28 x + 70 }{ 3 } \right) $$ $$ \left( 3 x^{4} + 4 x^{2} + 28 x + 5 \right) = \left( \frac{ 14 x^{2} - 28 x + 70 }{ 3 } \right) \cdot \color{magenta}{ \left( \frac{ 9 x^{2} + 18 x + 3 }{ 14 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 1}{3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 1}{3 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 9 x^{2} + 18 x + 3 }{ 14 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x^{2} + 6 x + 15 }{ 14 } \right) }{ \left( \frac{ 9 x^{2} + 18 x + 3 }{ 14 } \right) } $$ $$ \left( x^{2} + 2 x + 5 \right) \left( \frac{ 3}{14 } \right) - \left( 3 x^{2} + 6 x + 1 \right) \left( \frac{ 1}{14 } \right) = \left( 1 \right) $$ $$ \left( x^{4} + 6 x^{2} + 25 \right) = \left( x^{2} + 2 x + 5 \right) \cdot \color{magenta}{ \left( x^{2} - 2 x + 5 \right) } + \left( 0 \right) $$ $$ \left( 3 x^{4} + 4 x^{2} + 28 x + 5 \right) = \left( 3 x^{2} + 6 x + 1 \right) \cdot \color{magenta}{ \left( x^{2} - 2 x + 5 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} - 2 x + 5 \right) } $$ $$ \left( x^{4} + 6 x^{2} + 25 \right) \left( \frac{ 3}{14 } \right) - \left( 3 x^{4} + 4 x^{2} + 28 x + 5 \right) \left( \frac{ 1}{14 } \right) = \left( x^{2} - 2 x + 5 \right) $$
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