the polynomial 4x^3 +ax^2 +bx+9, where a and b consant, is denoted by f(x). when f(x) is divide by (x−2) the remainder is r and when divided by (x−3) the remaider is 6r. its further given that (x+3) is a factor of f(x). Show that b−a=14 and hence find a and b


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Question Number 155576 by MathsFan last updated on 02/Oct/21

$t h e p o l y n o m i a l 4 x^{3} + {a x}^{2} + b x + 9,$ $w h e r e a a n d b c o n s a n t, i s d e n o t e d b y$ $f (x) . w h e n f (x) i s d i v i d e b y (x - 2) t h e$ $r e m a i n d e r i s r a n d w h e n d i v i d e d b y$ $(x - 3) t h e r e m a i d e r i s 6 r . i t s f u r t h e r$ $g i v e n t h a t (x + 3) i s a f a c t o r o f f (x) .$ $S h o w t h a t b - a = 14$ $a n d h e n c e f i n d a a n d b$
Commented by Rasheed.Sindhi last updated on 02/Oct/21

$A r e y o u s u r e s i r / m a d a m t h a t$ $b - a = 14 ?$
Commented by MathsFan last updated on 02/Oct/21

$n o t r e a l l y s u r e s i r . i t r i e d b u t$ $d i d n t g e t b - a = 14 . t h a t s w h y i$ $p o s t e d i t o n t h i s p l a t f o r m f o r f u r t h e r$ $c o n v i c t i o n f r o m o t h e r s$
Commented by Rasheed.Sindhi last updated on 03/Oct/21

$I t h i n k b - a = - 25 i s c o r r e c t .$ $(a = 4, b = - 21)$
	Answered by Rasheed.Sindhi last updated on 02/Oct/21
	${ ((f(2)=4(2)^3 +a(2)^2 +b(2)+9=r)),((f(3)=4(3)^3 +a(3)^2 +b(3)+9=6r)),((f(−3)=4(−3)^3 +a(−3)^2 +b(−3)+9=0)) :} { ((32+4a+2b+9=r⇒4a+2b=r−41)),((108+9a+3b+9=6r⇒9a+3b=6r−117)),((−108+9a−3b+9=0⇒3a−b=33)) :} b=3a−33 4a+2b=r−41⇒4a+2(3a−33)=r−41 10a=r−41+66=r+25 r=10a−25 9a+3b=6r−117 ⇒9a+3(3a−33)=6(10a−25)−117 18a−99=60a−150−117 −42a=−168⇒a=((−168)/(−42))=4 b=3a−33=3(4)−33=−21 b=−21 b−a=−21−4=−25$
	${\begin{cases} f (2) = 4 {(2)}^{3} + a {(2)}^{2} + b (2) + 9 = r \\ f (3) = 4 {(3)}^{3} + a {(3)}^{2} + b (3) + 9 = 6 r \\ f (- 3) = 4 {(- 3)}^{3} + a {(- 3)}^{2} + b (- 3) + 9 = 0 \end{cases}$ ${\begin{cases} 32 + 4 a + 2 b + 9 = r \Rightarrow 4 a + 2 b = r - 41 \\ 108 + 9 a + 3 b + 9 = 6 r \Rightarrow 9 a + 3 b = 6 r - 117 \\ - 108 + 9 a - 3 b + 9 = 0 \Rightarrow 3 a - b = 33 \end{cases}$ $b = 3 a - 33$ $4 a + 2 b = r - 41 \Rightarrow 4 a + 2 (3 a - 33) = r - 41$ $10 a = r - 41 + 66 = r + 25$ $r = 10 a - 25$ $9 a + 3 b = 6 r - 117$ $\Rightarrow 9 a + 3 (3 a - 33) = 6 (10 a - 25) - 117$ $18 a - 99 = 60 a - 150 - 117$ $- 42 a = - 168 \Rightarrow a = \frac{- 168}{- 42} = 4$ $b = 3 a - 33 = 3 (4) - 33 = - 21$ $b = - 21$ $b - a = - 21 - 4 = - 25$
	Commented by MathsFan last updated on 02/Oct/21

	$a m a z i n g$
	Answered by Rasheed.Sindhi last updated on 02/Oct/21

	$\begin{array}{ccc} 2) & 4 & a & b & 9 \\ 8 & 2 a + 16 & 4 a + 2 b + 32 \\ 4 & a + 8 & 2 a + b + 16 & 4 a + 2 b + 41 = r . . . A \end{array}$ $\begin{array}{ccc} 3) & 4 & a & b & 9 \\ 12 & 3 a + 36 & 9 a + 3 b + 108 \\ 4 & a + 12 & 3 a + b + 36 & 9 a + 3 b + 117 = 6 r . . . B \end{array}$ $\begin{array}{ccc} - 3) & 4 & a & b & 9 \\ - 12 & - 3 a + 36 & 9 a - 3 b - 108 \\ 4 & a - 12 & - 3 a + b + 36 & 9 a - 3 b - 99 = 0 . . . C \end{array}$ $C \Rightarrow 3 a - b = 33 \Rightarrow b = 3 a - 33$ $B + C : 18 a + 18 = 6 r \Rightarrow r = 3 a + 3$ $A : 4 a + 2 b + 41 = r$ $4 a + 2 (3 a - 33) + 41 = 3 a + 3$ $10 a - 66 + 41 = 3 a + 3$ $7 a = 3 + 25 = 28$ $a = 4$ $b = 3 a - 33 = 3 (4) - 33 = - 21$ $b - a = - 21 - 4 = - 25$
	Commented by MathsFan last updated on 02/Oct/21

	$t h a n k y o u s i r$ $b u t w h i c h a p p r o a c h i s t h i s$
	Commented by Rasheed.Sindhi last updated on 02/Oct/21

	$T h i s i s s y n t h e t i c d i v i s i o n a p p r o a c h$
	Commented by peter frank last updated on 02/Oct/21

	$thanks$
	Commented by Tawa11 last updated on 03/Oct/21

	$Great sir$
	Commented by Rasheed.Sindhi last updated on 03/Oct/21

	$Y o u a r e c e r t a i n l y m i s s t a w a, a n$ $o l d f o r u m - f r i e n d ?$
	Answered by Rasheed.Sindhi last updated on 03/Oct/21

	$Verification :$ $a = 4, b = - 21$ $f (x) = 4 x^{3} + 4 x^{2} - 21 x + 9$ $∙ W h e n d i v i d e d b y x - 2$ $f (2) = 4 {(2)}^{3} + 4 {(2)}^{2} - 21 (2) + 9 = r$ $= 32 + 16 - 42 + 9 = 15 = r$ $∙ W h e n d i v i d e d b y x - 3$ $f (3) = 4 {(3)}^{3} + 4 {(3)}^{2} - 21 (3) + 9 = 6 r$ $= 108 + 36 - 63 + 9 = 90 = 6 \times 15 = 6 r$ $∙ x + 3 i s f a c t o r o f f (x)$ $\begin{array}{ccc} - 3) & 4 & 4 & - 21 & 9 \\ - 12 & 24 & - 9 \\ 4 & - 8 & 3 & 0 \end{array}$ $4 x^{3} + 4 x^{2} - 21 x + 9$ $= (x + 3) (4 x^{2} - 8 x + 3)$ $V e r i f i e d$
	Answered by Rasheed.Sindhi last updated on 04/Oct/21

	$A n O ther W ay$ $^{∙} x + 3 is factor of 4 x^{3} + {a x}^{2} + b x + 9 .$ $L e t o t h e r f a c t o r i s 4 x^{2} + p x + 3$ $∴ f (x) = (x + 3) (4 x^{2} + p x + 3)$ $= 4 x^{3} + (p + 12) x^{2} + (3 p + 3) x + 9$ $C o m p a r i n g c o e f f i c i e n t s$ $a = p + 12 \Rightarrow p = a - 12,$ $b = 3 p + 3 = 3 (a - 12) + 3 = 3 a - 33$ $∴ f (x) = 4 x^{3} + {a x}^{2} + (3 a - 33) x + 9$ $^{∙} f (x) d i v i d e d b y x - 2 l e a v i n g$ $r e m a i n d e r r$ $f (2) = 4 {(2)}^{3} + a {(2)}^{2} + (3 a - 33) (2) + 9 = r$ $= 32 + 4 a + 6 a - 66 + 9 = r$ $10 a - 25 = r$ $^{∙} f (x) d i v i d e d b y x - 3 l e a v i n g$ $r e m a i n d e r 6 r$ $f (3) = 4 {(3)}^{3} + a {(3)}^{2} + (3 a - 33) (3) + 9 = 6 r$ $= 108 + 9 a + 9 a - 99 + 9 = 6 (10 a - 25)$ $18 a + 18 = 60 a - 150$ $42 a = 168 \Rightarrow a = 4$ $b = 3 a - 33 = 3 (4) - 33 = - 21$ $b = - 21$ $b - a = - 21 - 4 = - 25$
	Answered by ajfour last updated on 03/Oct/21

	$x = - 3 \Rightarrow$ $4 (- 27) + 9 a - 3 b = - 9$ $\Rightarrow 9 a - 3 b = 99 . . (i)$ $f (2) = r \Rightarrow$ $32 + 4 a + 2 b + 9 = r$ $f (3) = 6 r \Rightarrow$ $4 (27) + 9 a + 3 b + 9 = 6 r$ $3 a + b + 39 = 2 r = 64 + 8 a + 4 b + 18$ $\Rightarrow 5 a + 3 b = - 43 . . (i i)$ $(i) + (i i) \Rightarrow$ $14 a = 56 \Rightarrow a = 4$ $b = 3 a - 33 = - 21$ $b - a = - 25$
	Answered by Rasheed.Sindhi last updated on 04/Oct/21
	$▶ {: ((f(3)=6r)),((f(2)=r⇒6f(2)=6r)) }⇒f(3)−6f(2)=0 ⇒(4.3^3 +a.3^2 +b.3+9) −6(4.2^3 +a.2^2 +b.2+9)=0 ⇒15a+9b=−129.............A ▶f(−3)=4(−3)^3 +a(−3)^2 +b(−3)+9=0 −108+9a−3b+9=0 9a−3b=99 27a−9b=297.............B A+B: 42a=168⇒a=((168)/(42))=4 B:15(4)+9b=−129⇒b=((−129−60)/9)=−21 a=4 & b=−21 b−a=−21−4=−25$
	$▸ \begin{matrix} f (3) = 6 r \\ f (2) = r \Rightarrow 6 f (2) = 6 r \end{matrix}} \Rightarrow f (3) - 6 f (2) = 0$ $\Rightarrow (4 . 3^{3} + a . 3^{2} + b . 3 + 9)$ $- 6 (4 . 2^{3} + a . 2^{2} + b . 2 + 9) = 0$ $\Rightarrow 15 a + 9 b = - 129 . . . . . . . . . . . . . A$ $▸ f (- 3) = 4 {(- 3)}^{3} + a {(- 3)}^{2} + b (- 3) + 9 = 0$ $- 108 + 9 a - 3 b + 9 = 0$ $9 a - 3 b = 99$ $27 a - 9 b = 297 . . . . . . . . . . . . . B$ $A + B : 42 a = 168 \Rightarrow a = \frac{168}{42} = 4$ $B : 15 (4) + 9 b = - 129 \Rightarrow b = \frac{- 129 - 60}{9} = - 21$ $a = 4 & b = - 21$ $b - a = - 21 - 4 = - 25$
	Answered by Rasheed.Sindhi last updated on 04/Oct/21

	$\underset{A}{\underset{⏟}{f (3) = 6 r}} \land \underset{B}{\underset{⏟}{f (2) = r}}$ $\frac{A}{B} \Rightarrow \frac{f (3)}{f (2)} = 6 \Rightarrow f (3) - 6 f (2) = 0$ $\Rightarrow 4 . 3^{3} + a . 3^{2} + b . 3 + 9 - 6 (4 . 2^{3} + a . 2^{2} + b . 2 + 9) = 0$ $\Rightarrow 4 . 3^{3} + a . 3^{2} + b . 3 + 9 - 24 . 2^{3} - 6 a . 2^{2} - 6 b . 2 - 54 = 0$ $\Rightarrow 108 + 9 a + 3 b + 9 - 192 - 24 a - 12 b - 54 = 0$ $- 15 a - 9 b = 129$ $5 a + 3 b = - 43 . . . . . . . . . . . . . . . (i)$ $▸ f (- 3) = 4 {(- 3)}^{3} + a {(- 3)}^{2} + b (- 3) + 9 = 0$ $\Rightarrow - 108 + 9 a - 3 b + 9 = 0$ $\Rightarrow 3 a - b = 33 . . . . . . . . . . . . . . . . (i i)$ $(i) & (i i) : a = 4, b = - 21$ $b - a = - 21 - 4 = - 25$
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