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Question

(A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $p(x)$,then $p(a) = 0$.Here,the given factor is $(x-1)$,so we must have $p(

Vedclass · Accepted Answer

$p(x)=x^{2}+2x-3$

Vedclass · Answer

(A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $p(x)$,then $p(a) = 0$.Here,the given factor is $(x-1)$,so we must have $p(1) = 0$.Let us check each option:$A) p(1) = (1)^{2} + 2(1) - 3 = 1 + 2 - 3 = 0$. Since $p(1) = 0$,$(x-1)$ is a factor.$B) p(1) = (1)^{2} + 4(1) + 3 = 1 + 4 + 3 = 8 
eq 0$.$C) p(1) = (1)^{2} + 5(1) + 6 = 1 + 5 + 6 = 12 
eq 0$.$D) p(1) = (1)^{2} + (1) - 6 = 1 + 1 - 6 = -4 
eq 0$.Therefore,the correct option is $A$.

One of the factors of $\ldots \ldots \ldots \ldots$ is $(x-1).$

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