If (x-4) is the G.C.D. of x^{2}-x-12 and x^{2 | vedclass

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(C) The $G.C.D.$ (Greatest Common Divisor) is given as $(x-4)$.Let $p(x) = x^{2}-x-12$. Factoring this,we get $p(x) = (x-4)(x+3)$.Let $q(x) = x^{2}-mx

Vedclass · Accepted Answer

$2$

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(C) The $G.C.D.$ (Greatest Common Divisor) is given as $(x-4)$.Let $p(x) = x^{2}-x-12$. Factoring this,we get $p(x) = (x-4)(x+3)$.Let $q(x) = x^{2}-mx-8$.Since $(x-4)$ is the $G.C.D.$,it must be a factor of $q(x)$.According to the Factor Theorem,if $(x-4)$ is a factor of $q(x)$,then $q(4) = 0$.Substituting $x = 4$ into $q(x)$:$q(4) = (4)^{2} - m(4) - 8 = 0$$16 - 4m - 8 = 0$$8 - 4m = 0$$4m = 8$$m = 2$.

If $(x-4)$ is the $G.C.D.$ of $x^{2}-x-12$ and $x^{2}-mx-8$,find the value of $m$.

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