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Let $P(x) = x^5 + a_1x^4 + a_2x^3+ a_3x^2+ a_4x + 14$

and $Q(x) = x^5 + b_1x^4 + b_2x^3+ b_3x^2+ b_4x + 42$

be polynomials with integral coefficients.

If $P(x)$ and $Q(x)$ have four distinct common rational roots.

Find all possible $Q(x)$.

Please check my answer.

$P(x) = (x-1)(x+1)(x+2)(x+7)(x-1)\Rightarrow Q(x) = (x-1)(x+1)(x+2)(x+7)(x-3)$

$P(x) = (x-1)(x+1)(x-2)(x-7)(x-1)\Rightarrow Q(x) = (x-1)(x+1)(x-2)(x-7)(x-3)$

$P(x) = (x-1)(x+1)(x-2)(x+7)(x+1)\Rightarrow Q(x) = (x-1)(x+1)(x-2)(x+7)(x+3)$

$P(x) = (x-1)(x+1)(x+2)(x-7)(x+1)\Rightarrow Q(x) = (x-1)(x+1)(x+2)(x-7)(x+3)$

user403160
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1 Answers1

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Since the product of the roots of $P$ is an integer, if $P$ has four different rational roots, it has five rational roots. The same holds for $Q$.

By the rational root theorem, all roots of $P$ and $Q$ are integers and:

  • the roots of $P$ are integers that divide $14$,

  • the roots of $Q$ are integers that divide $42$.

Moreover, the product of the roots of $P$ is $14$ and the product of the roots of $Q$ is $42$. Therefore:

  • for $P$, exactly one root is $\pm 2$ and exactly one root is $\pm 7$. The others are $\pm 1$.

  • for $Q$, exactly one root is $\pm 2$, exactly one root is $\pm 3$, and exactly one root is $\pm 7$. The others are $\pm 1$.

The possible combinations follow from this.

lhf
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