If 1 and 3 are the zeros of the polynomial p( | vedclass

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(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of

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$-3, \frac{5}{2}$

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(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of $p(x)$.Dividing $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$ by $(x^2 - 4x + 3)$:$2x^4 - 7x^3 - 13x^2 + 63x - 45 = (x^2 - 4x + 3)(2x^2 + x - 15)$.Now,factorize the quadratic $2x^2 + x - 15$:$2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)$.Setting these factors to zero,we get $x = -3$ and $x = \frac{5}{2}$.Thus,the remaining zeros are $-3$ and $\frac{5}{2}$.

If $1$ and $3$ are the zeros of the polynomial $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$,then find the remaining zeros of $p(x)$.

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