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Question Number 115117 by bobhans last updated on 23/Sep/20

What is the value of a and b when   3x^4 +6x^3 −ax^2 −bx−12 is completely  divisible by x^2 −3 ?

$${What}\:{is}\:{the}\:{value}\:{of}\:{a}\:{and}\:{b}\:{when}\: \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} −{ax}^{\mathrm{2}} −{bx}−\mathrm{12}\:{is}\:{completely} \\ $$$${divisible}\:{by}\:{x}^{\mathrm{2}} −\mathrm{3}\:? \\ $$

Answered by john santu last updated on 23/Sep/20

Write your polynomial as   p(x)= x^2 (3x^2 +6x−a)−bx−12  now computing the remainder of  the division is easy. Substituting   3 for x^2  ⇒ 3(3.3+6x−a)−bx−12 = 0  ⇒ 3(6x+9−a)−bx−12 = 0x + 0   { ((18−b=0→b=18)),((27−3a−12=0→a= 5  )) :}

$${Write}\:{your}\:{polynomial}\:{as}\: \\ $$$${p}\left({x}\right)=\:{x}^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−{a}\right)−{bx}−\mathrm{12} \\ $$$${now}\:{computing}\:{the}\:{remainder}\:{of} \\ $$$${the}\:{division}\:{is}\:{easy}.\:{Substituting}\: \\ $$$$\mathrm{3}\:{for}\:{x}^{\mathrm{2}} \:\Rightarrow\:\mathrm{3}\left(\mathrm{3}.\mathrm{3}+\mathrm{6}{x}−{a}\right)−{bx}−\mathrm{12}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{3}\left(\mathrm{6}{x}+\mathrm{9}−{a}\right)−{bx}−\mathrm{12}\:=\:\mathrm{0}{x}\:+\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{18}−{b}=\mathrm{0}\rightarrow{b}=\mathrm{18}}\\{\mathrm{27}−\mathrm{3}{a}−\mathrm{12}=\mathrm{0}\rightarrow{a}=\:\mathrm{5}\:\:}\end{cases} \\ $$

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