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


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

$W h a t i s t h e v a l u e o f a a n d b w h e n$ $3 x^{4} + 6 x^{3} - {a x}^{2} - b x - 12 i s c o m p l e t e l y$ $d i v i s i b l e b y x^{2} - 3 ?$
	Answered by john santu last updated on 23/Sep/20
	$Write your polynomial as p(x)= x^2 (3x^2 +6x−a)−bx−12 now computing the remainder of the division is easy. Substituting 3 for x^2 ⇒ 3(3.3+6x−a)−bx−12 = 0 ⇒ 3(6x+9−a)−bx−12 = 0x + 0 { ((18−b=0→b=18)),((27−3a−12=0→a= 5 )) :}$
	$W r i t e y o u r p o l y n o m i a l a s$ $p (x) = x^{2} (3 x^{2} + 6 x - a) - b x - 12$ $n o w c o m p u t i n g t h e r e m a i n d e r o f$ $t h e d i v i s i o n i s e a s y . S u b s t i t u t i n g$ $3 f o r x^{2} \Rightarrow 3 (3.3 + 6 x - a) - b x - 12 = 0$ $\Rightarrow 3 (6 x + 9 - a) - b x - 12 = 0 x + 0$ ${\begin{cases} 18 - b = 0 \to b = 18 \\ 27 - 3 a - 12 = 0 \to a = 5 \end{cases}$
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