![Let\ p(x)\ be\ x^3+kx^2+7x+5\\Let\ g(x)\ be\ (x+6).\\Hence, \ \\The\ zero\ of\ g(x) :\\(x+6)=0\\x=-6\\\\Hence,\\p(x)\ =\ x^3+kx^2+7x+5\\p(-6)=(-6)^3+k(-6)^2+7*-6+5\\=-216+36k-42+5\\=-253+36k\\\\Through\ Remainder\ Theorem\ we\ get\ that\ R=-1\ &\ R=-253+36k [Given]\\Hence,\\-253+36k=-1\\36k=-1+253\\36k=252\\k=7](/tpl/images/0921/8467/b3099.png)
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