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Question Number 120355 by TITA last updated on 30/Oct/20

Commented by TITA last updated on 30/Oct/20

$p l e a s e h e l p$
Commented by TITA last updated on 30/Oct/20

$p l e a s e h e l p$
	Answered by john santu last updated on 30/Oct/20

	$(1) f (2) = 4$ $(2) f (- 3) = - 1$ $(3) f (x) : (x + 3) (x - 2)$ $\Rightarrow r (x) = \frac{x + 3}{2 + 3} f (2) + \frac{x - 2}{- 3 - 2} f (- 3)$ $\Rightarrow r (x) = \frac{4}{5} (x + 3) + \frac{1}{5} (x - 2)$ $\Rightarrow r (x) = \frac{4 x + 12 + x - 2}{5} = \frac{5 x + 10}{5}$ $\Rightarrow r (x) = x + 2$
	Commented by TITA last updated on 30/Oct/20

	$t h a n k s i r$
	Answered by benjo_mathlover last updated on 30/Oct/20
	$in other way let f(x) = g(x)(x+3)(x−2)+px+q where px+q is remainder when f(x) divided by x^2 +x−6 for x=2⇒2p+q=4 for x=−3⇒−3p+q=−1 ((( 2 1)),((−3 1)) ) ((p),(q) ) = ((( 4)),((−1)) ) → { ((p=( determinant ((( 4 1)),((−1 1)))/5)= (5/5)=1)),((q=( determinant ((( 2 4)),((−3 −1)))/5)=((10)/5)=2)) :} thus r(x)=1.x+2=x+2$
	$i n o t h e r w a y$ $l e t f (x) = g (x) (x + 3) (x - 2) + p x + q$ $w h e r e p x + q i s r e m a i n d e r w h e n f (x) d i v i d e d$ $b y x^{2} + x - 6$ $f o r x = 2 \Rightarrow 2 p + q = 4$ $f o r x = - 3 \Rightarrow - 3 p + q = - 1$ $(\begin{matrix} 2 1 \\ - 3 1 \end{matrix}) (\begin{matrix} p \\ q \end{matrix}) = (\begin{matrix} 4 \\ - 1 \end{matrix})$ $\to {\begin{cases} p = \frac{\| \begin{array}{c} 4 1 \\ - 1 1 \end{array} \|}{5} = \frac{5}{5} = 1 \\ q = \frac{\| \begin{array}{c} 2 4 \\ - 3 - 1 \end{array} \|}{5} = \frac{10}{5} = 2 \end{cases}$ $t h u s r (x) = 1 . x + 2 = x + 2$
	Answered by mathmax by abdo last updated on 30/Oct/20
	$f(x)=(x−2)q(x)+4 and f(x)=(x+3)v(x)−1 ⇒ f(2)=4 and f(−3)=−1 f(x)=h(x)(x^2 +x−6)+ax+b we have f(2) =2a+b=4 and f(−3)=−3a+b =−1 ⇒ { ((2a+b=4)),((3a−b=1 ⇒ { ((5a=5)),((b=4−2a ⇒ { ((a=1)),((b=2 ⇒)) :})) :})) :} f(x)=h(x^2 +x−6)+x+2 ⇒r(x)=x+2$
	$f (x) = (x - 2) q (x) + 4 and f (x) = (x + 3) v (x) - 1 \Rightarrow$ $f (2) = 4 and f (- 3) = - 1$ $f (x) = h (x) (x^{2} + x - 6) + ax + b$ $we have f (2) = 2 a + b = 4 and f (- 3) = - 3 a + b = - 1 \Rightarrow$ ${\begin{cases} 2 a + b = 4 \\ 3 a - b = 1 \Rightarrow {\begin{cases} 5 a = 5 \\ b = 4 - 2 a \Rightarrow {\begin{cases} a = 1 \\ b = 2 \Rightarrow \end{cases} \end{cases} \end{cases}$ $f (x) = h (x^{2} + x - 6) + x + 2 \Rightarrow r (x) = x + 2$
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