If x^2-3x+2 is a factor of x^4-ax^2+b,then th | vedclass

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(B) Given that $x^2-3x+2$ is a factor of $P(x) = x^4-ax^2+b$. Factorizing the divisor: $x^2-3x+2 = (x-1)(x-2)$. Since $(x-1)$ and $(x-2)$ are factors,

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$x^2-9x+20=0$

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(B) Given that $x^2-3x+2$ is a factor of $P(x) = x^4-ax^2+b$. Factorizing the divisor: $x^2-3x+2 = (x-1)(x-2)$. Since $(x-1)$ and $(x-2)$ are factors,$P(1) = 0$ and $P(2) = 0$. For $x=1$: $(1)^4 - a(1)^2 + b = 0 \implies 1 - a + b = 0 \implies -a + b = -1$ (Equation $i$). For $x=2$: $(2)^4 - a(2)^2 + b = 0 \implies 16 - 4a + b = 0 \implies -4a + b = -16$ (Equation $ii$). Subtracting (Equation $ii$) from (Equation $i$): $(-a+b) - (-4a+b) = -1 - (-16) \implies 3a = 15 \implies a = 5$. Substituting $a=5$ into (Equation $i$): $-5 + b = -1 \implies b = 4$. The roots of the required quadratic equation are $a=5$ and $b=4$. The equation is given by $x^2 - (a+b)x + ab = 0$. Substituting the values: $x^2 - (5+4)x + (5 	imes 4) = 0 \implies x^2 - 9x + 20 = 0$.

If $x^2-3x+2$ is a factor of $x^4-ax^2+b$,then the equation whose roots are $a$ and $b$ is

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