If x^2 - 3x + 2 is a factor of x^4 - px^2 + q | vedclass

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(D) Let $f(x) = x^4 - px^2 + q$. Since $x^2 - 3x + 2$ is a factor of $f(x)$,the roots of $x^2 - 3x + 2 = 0$ must also be roots of $f(x) = 0$.$x^2 - 3x

Vedclass · Accepted Answer

$(5, 4)$

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(D) Let $f(x) = x^4 - px^2 + q$. Since $x^2 - 3x + 2$ is a factor of $f(x)$,the roots of $x^2 - 3x + 2 = 0$ must also be roots of $f(x) = 0$.$x^2 - 3x + 2 = (x - 1)(x - 2) = 0$,so the roots are $x = 1$ and $x = 2$.For $x = 1$: $1^4 - p(1)^2 + q = 0$ $\Rightarrow 1 - p + q = 0$ $\Rightarrow p - q = 1$ (Equation $i$).For $x = 2$: $2^4 - p(2)^2 + q = 0$ $\Rightarrow 16 - 4p + q = 0$ $\Rightarrow 4p - q = 16$ (Equation $ii$).Subtracting $(i)$ from (ii): $(4p - q) - (p - q) = 16 - 1$ $\Rightarrow 3p = 15$ $\Rightarrow p = 5$.Substituting $p = 5$ into $(i)$: $5 - q = 1 \Rightarrow q = 4$.Thus,$(p, q) = (5, 4)$.

If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 + q$,then $(p, q) = $

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