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Question Number 136693 by liberty last updated on 25/Mar/21

Commented by Olaf last updated on 25/Mar/21
$f(x) = mx^3 +5x^2 +kx−18 = (6−x−x^2 )(−mx−3) =(x^2 +x−6)(mx+3) = mx^3 +(3+m)x^2 +(3−6m)x−18 By identification : { ((3+m = 5 (1))),((3−6m = k (2))) :} (1) : m = 2 (2) : k = 3−6m = −9 then f(x) = 2x^3 +5x^2 −9x−18 = (6−x−x^2 )(−2x−3)$
$f (x) = {m x}^{3} + 5 x^{2} + k x - 18$ $= (6 - x - x^{2}) (- m x - 3)$ $= (x^{2} + x - 6) (m x + 3)$ $= {m x}^{3} + (3 + m) x^{2} + (3 - 6 m) x - 18$ $By identification :$ ${\begin{cases} 3 + m = 5 (1) \\ 3 - 6 m = k (2) \end{cases}$ $(1) : m = 2$ $(2) : k = 3 - 6 m = - 9$ $then f (x) = 2 x^{3} + 5 x^{2} - 9 x - 18$ $= (6 - x - x^{2}) (- 2 x - 3)$
	Answered by Rasheed.Sindhi last updated on 25/Mar/21

	$6 - x - x^{2} = - (x + 3) (x - 2)$ $∴ x + 3 & x - 2 a r e a l s o f a c t o r s o f f (x)$ $B y s y n t h e t i c d i v i s i o n$ $\begin{array}{ccccc} 2) & m & 5 & k & - 18 \\ 2 m & 4 m + 10 & 8 m + 2 k + 20 \\ - 3) & m & 2 m + 5 & 4 m + k + 10 & \underset{4 m + k = - 1}{8 m + 2 k + 2 = 0} \\ - 3 m & 3 m - 15 \\ m & - m + 5 & 7 m + k - 5 = 0 \end{array}$ $4 m + k = - 1$ $7 m + k = 5$ $3 m = 6 \Rightarrow m = 2$ $k = 5 - 7 m = 5 - 7 (2) = - 9$
	Answered by bramlexs22 last updated on 25/Mar/21
	$method(1) by Remainder theorem consider 6−x−x^2 = 0 or x^2 +x−6=0 we get (x+3)(x−2)=0⇒x=−3, 2 since 6−x−x^2 is a factor of polynomes f(x) so f(−3) and f(2) is zeros to f(x). f(2)=8m+20+2k−18=0 8m+2k=−2→4m+k=−1...(i) f(−3)=−27m+45−3k−18=0 −27m−3k=−27→9m+k=9...(ii) eliminate (i)&(ii) gives 5m = 10 { ((m=2)),((k=−9)) :}$
	$m e t h o d (1)$ $b y R e m a i n d e r t h e o r e m$ $c o n s i d e r 6 - x - x^{2} = 0$ $o r x^{2} + x - 6 = 0 w e g e t$ $(x + 3) (x - 2) = 0 \Rightarrow x = - 3, 2$ $s i n c e 6 - x - x^{2} i s a f a c t o r o f$ $p o l y n o m e s f (x) s o f (- 3) a n d$ $f (2) i s z e r o s t o f (x) .$ $f (2) = 8 m + 20 + 2 k - 18 = 0$ $8 m + 2 k = - 2 \to 4 m + k = - 1 . . . (i)$ $f (- 3) = - 27 m + 45 - 3 k - 18 = 0$ $- 27 m - 3 k = - 27 \to 9 m + k = 9 . . . (i i)$ $e l i m i n a t e (i) & (i i) g i v e s$ $5 m = 10 {\begin{cases} m = 2 \\ k = - 9 \end{cases}$
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