Question

mr kitts works at a music store last week she sold 6 more than 3 times the number of CDs that she sold this week. mr. kitts sold a total of 110 CDs she sold last week, and t the number of CDs she sold this week?

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow p(x)= 4-2x^3+3x^2-ax+3a-7$

$\sf\dashrightarrow g(x)=x+1$

$\sf\dashrightarrow when,\:p(x)\:gets\:divided\:by\:(x+1),\:it\:leaves \:the\:remainder\:is\:19$

$\sf\therefore r(x)=19$

$\sf\dashrightarrow another\:g(x)= (x+2)$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow THE\:VALUE\:OF\:x$

$\sf\dashrightarrow AND,THE\:REMAINDER\:WHEN\:P(X)\:IS\:DIVIDED \:BY\:(x+2)$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore r(x)=19$

$\sf\implies 4-2x^3+3x^2-ax+3a-7=19$

$\sf\therefore g(x)=x+1$

$\sf\implies x=-1$

$\sf\implies 2(-1)^3+3(-1)^2-a(-1)+3a-7+4=19$

$\sf\implies 2+3+a+3a-3=19$

$\sf\implies 5+4a-3=19$

$\sf\implies 4a=19-2$

$\sf\implies 4a=17$

$\sf\implies a =\dfrac{17}{4}$

$\large{\boxed{\bf{ \star\:\: a= \dfrac{17}{4}\:\: \star}}}$

_______________

$\sf\dashrightarrow NOW, FINDING\: FOR\:g(x)= (x+2).$

$\sf\therefore x+2=0$

$\sf\implies x=-2$

$\sf\therefore a= \dfrac{17}{4}$

$\sf\therefore p(x)= 4-2x^3+3x^2-ax+3a-7$

$\sf\implies 2x^3+3x^2-ax+3a-3$

$\sf\implies 2(-2)^3+3(-2)^2-\bigg( \dfrac{17}{4} \bigg) (-2)+3\bigg( \dfrac{17}{4}-3$

$\sf\implies 2(-2)^3+3(-2)^2-\bigg( \dfrac{17}{\cancel{4}} \bigg) (\cancel{-2})+3\bigg( \dfrac{17}{4}\bigg)-3$

$\sf\implies -16+12+ \dfrac{17}{2}+3\times \bigg(\dfrac{17}{4} \bigg) - 3$

$\sf\implies \left( -8 +\dfrac{17}{2} \right) + \left( \dfrac{51}{4}-3\right)$

$\sf\implies \left( \dfrac{-16+17}{2} \right) + \left( \dfrac{51-12}{4}\right)$

$\sf\implies \dfrac{1}{2}+ \dfrac{39}{4}$

$\sf\implies [\dfrac{4+78}{4}/tex]</p><p>[tex]\sf\implies \dfrac{82}{8}$

$\sf\implies \cancel\dfrac{82}{8}$

$\sf\implies \dfrac{41}{4}$

$\large{\boxed{\bf{ \star\:\:remainder\:of\:p(x)= \dfrac{41}{4} \:\: \star}}}$

________________

Answer 2

$\large\underline{\underline{\bold{\pink{\mathfrak{AnSwEr}}}}}$

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow p(x)= 4-2x^3+3x^2-ax+3a-7$

$\sf\dashrightarrow g(x)=x+1$

$\sf\dashrightarrow when,\:p(x)\:gets\:divided\:by\:(x+1),\:it\:leaves \:the\:remainder\:is\:19$

$\sf\therefore r(x)=19$

$\sf\dashrightarrow another\:g(x)= (x+2)$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow THE\:VALUE\:OF\:x$

$\sf\dashrightarrow AND,THE\:REMAINDER\:WHEN\:P(X)\:IS\:DIVIDED \:BY\:(x+2)$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore r(x)=19$

$\sf\implies 4-2x^3+3x^2-ax+3a-7=19$

$\sf\therefore g(x)=x+1$

$\sf\implies x=-1$

$\sf\implies 2(-1)^3+3(-1)^2-a(-1)+3a-7+4=19$

$\sf\implies 2+3+a+3a-3=19$

$\sf\implies 5+4a-3=19$

$\sf\implies 4a=19-2$

$\sf\implies 4a=17$

$\sf\implies a =\dfrac{17}{4}$

$\large{\boxed{\bf{ \star\:\: a= \dfrac{17}{4}\:\: \star}}}$

$\sf\dashrightarrow NOW, FINDING\: FOR\:g(x)= (x+2).$

$\sf\therefore x+2=0$

$\sf\implies x=-2$

$\sf\therefore a= \dfrac{17}{4}$

$\sf\therefore p(x)= 4-2x^3+3x^2-ax+3a-7$

$\sf\implies 2x^3+3x^2-ax+3a-3$

$\sf\implies 2(-2)^3+3(-2)^2-\bigg( \dfrac{17}{4} \bigg) (-2)+3\bigg( \dfrac{17}{4}-3$

$\sf\implies 2(-2)^3+3(-2)^2-\bigg( \dfrac{17}{\cancel{4}} \bigg) (\cancel{-2})+3\bigg( \dfrac{17}{4}\bigg)-3$

$\sf\implies -16+12+ \dfrac{17}{2}+3\times \bigg(\dfrac{17}{4} \bigg) - 3$

$\sf\implies \left( -8 +\dfrac{17}{2} \right) + \left( \dfrac{51}{4}-3\right)$

$\sf\implies \left( \dfrac{-16+17}{2} \right) + \left( \dfrac{51-12}{4}\right)$

$\sf\implies \dfrac{1}{2}+ \dfrac{39}{4}$

$\sf\implies [\dfrac{4+78}{4}/tex]</p><p>[tex]\sf\implies \dfrac{82}{8}$

$\sf\implies \cancel\dfrac{82}{8}$

$\sf\implies \dfrac{41}{4}$

$\large{\boxed{\bf{ \star\:\:remainder\:of\:p(x)= \dfrac{41}{4} \:\: \star}}}$