If x^2-5x+6 is a factor of f(x)=x^4-17x^3+kx^ | vedclass

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(B) Given that $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$. We can factorize $x^2-5x+6$ as $(x-2)(x-3)$. Since $(x-2)$ and $(x-3)$ are fa

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$x^2-12x+35$

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(B) Given that $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$. We can factorize $x^2-5x+6$ as $(x-2)(x-3)$. Since $(x-2)$ and $(x-3)$ are factors of $f(x)$,we have $f(2)=0$ and $f(3)=0$. Let the other quadratic factor be $x^2+ax+b$. Since the leading coefficient of $f(x)$ is $1$ and the constant term is $210$,we have $(x^2-5x+6)(x^2+ax+35) = x^4-17x^3+kx^2-247x+210$. Comparing the coefficient of $x^3$: $a-5 = -17$,which gives $a = -12$. Thus,the other quadratic factor is $x^2-12x+35$.

If $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$,then the other quadratic factor of $f(x)$ is

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