If f(x)=2 x^4-13 x^2+a x+b is divisible by x^ | vedclass

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(C) Given,$f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2 = (x-2)(x-1)$.Since $f(x)$ is divisible by $(x-2)$ and $(x-1)$,we must have $f(2)=0$ an

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$(9, 2)$

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(C) Given,$f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2 = (x-2)(x-1)$.Since $f(x)$ is divisible by $(x-2)$ and $(x-1)$,we must have $f(2)=0$ and $f(1)=0$.For $f(2)=0$:$2(2)^4-13(2)^2+a(2)+b=0$$2(16)-13(4)+2a+b=0$$32-52+2a+b=0$$2a+b=20$ ... $(i)$For $f(1)=0$:$2(1)^4-13(1)^2+a(1)+b=0$$2-13+a+b=0$$a+b=11$ ... $(ii)$Subtracting Eq. $(ii)$ from Eq. $(i)$:$(2a+b)-(a+b)=20-11$$a=9$Substituting $a=9$ in Eq. $(ii)$:$9+b=11$$b=2$Thus,$(a, b) = (9, 2)$.

If $f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2$,then $(a, b)$ is equal to

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