Please can someone help with a simplier method of solving this question Q1; Given that the expression x^3 +x^2 −4x +5 and x^3 +3x−7 leave same remainder when divided by (x−a) find the possible values of a


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Question Number 33455 by Rio Mike last updated on 16/Apr/18

$P l e a s e c a n s o m e o n e h e l p w i t h a$ $s i m p l i e r m e t h o d o f s o l v i n g t h i s q u e s t i o n$ $Q 1;$ $G i v e n t h a t t h e e x p r e s s i o n x^{3} + x^{2} - 4 x + 5$ $a n d x^{3} + 3 x - 7 l e a v e s a m e r e m a i n d e r$ $w h e n d i v i d e d b y (x - a) f i n d t h e p o s s i b l e$ $v a l u e s o f a$
Commented by prof Abdo imad last updated on 17/Apr/18

$x^{3} + x^{2} - 4 x + 5 = (x - a) Q (x) + r (x)$ $x^{3} + 3 x - 7 = (x - a) V (x) + r (x) \Rightarrow$ $r (a) = a^{3} + a^{2} - 4 a + 5 = a^{3} + 3 a - 7 \Rightarrow$ $a^{2} - 4 a + 5 - 3 a + 7 = 0 \Rightarrow a^{2} - 7 a + 12 = 0$ $Δ = 49 - 48 = 1 \Rightarrow a_{1} = \frac{7 + 1}{2} = 4$ $a_{2} = \frac{7 - 1}{2} = 3 .$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Apr/18

$a^{3} + a^{2 \overset{}{}} - 4 a + 5 = a^{3} + 3 a - 7$ $a^{2} - 7 a + 12 = 0$ $(a - 3) (a - 4) = 0$ $a = 3 o r a = 4$
	Answered by ajfour last updated on 16/Apr/18

	$l e t r e m a i n d e r = R$ $\Rightarrow a^{3} + a^{2} - 4 a + 5 = a^{3} + 3 a - 7 = R$ $\Rightarrow a^{2} - 7 a + 12 = 0$ $o r (a - 3) (a - 4) = 0$ $\Rightarrow a = 3, 4 .$
	Commented by Rio Mike last updated on 16/Apr/18

	$t h a n k s v e r y s i m p l e n o w$
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