P(x) is polynomial P(x) = ((x^4 + 2ax^3 − bx − 5)/((x + 1)^2 )) Find: b = ?


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Question Number 208362 by hardmath last updated on 13/Jun/24

$P (x) is polynomial$ $P (x) = \frac{x^{4} + {2 ax}^{3} - bx - 5}{{(x + 1)}^{2}}$ $Find : b = ?$
	Answered by mr W last updated on 13/Jun/24

	$x^{4} + 2 {a x}^{3} - b x - 5 = {(x + 1)}^{2} (x^{2} + p x + q)$ $x^{4} + 2 {a x}^{3} - b x - 5 = x^{4} + (2 + p) x^{3} + (1 + q + 2 p) x^{2} + (p + 2 q) x + q$ $q = - 5$ $p + 2 q = - b \Rightarrow b = 8 ✓$ $1 + q + 2 p = 0 \Rightarrow p = 2$ $2 + p = 2 a \Rightarrow a = 2$
	Answered by efronzo1 last updated on 13/Jun/24

	$\begin{array}{cccc} 1 & 2 a & 0 & - b & - 5 \\ - 1 & * & * & - 1 & 2 - 2 a & 4 a - 3 \\ - 2 & * & - 2 & 4 - 4 a & 8 a - 6 & * \\ 1 & 2 a - 2 & 3 - 4 a & 6 a - 4 - b & 4 a - 8 \end{array}$ $\Rightarrow 4 a - 8 = 0 \Rightarrow a = 2$ $\Rightarrow 6 (2) - 4 = b \Rightarrow b = 8$
	Answered by Rasheed.Sindhi last updated on 14/Jun/24

	$x^{4} + 2 {a x}^{3} - b x - 5 i s d i v i s i b l e b y x + 1$ $s o b y s y n t h e t i c d i v i s i o n :$ $\begin{array}{ccc} - 1) & 1 & 2 a & 0 & - b & - 5 \\ - 1 & - 2 a + 1 & 2 a - 1 & - 2 a + b + 1 \\ 1 & 2 a - 1 & - 2 a + 1 & 2 a - b - 1 & - 2 a + b - 4 = 0 \end{array}$ $Q (x) = x^{3} + (2 a - 1) x^{2} + (- 2 a + 1) x + (2 a - b - 1)$ $T h e q u o t i e n t i s a g a i n d i v i s i b l e b y x + 1$ $s o a g a i n b y s y n t h e t i c d i v i s i o n :$ $\begin{array}{ccc} - 1) & 1 & 2 a - 1 & - 2 a + 1 & 2 a - b - 1 \\ - 1 & - 2 a + 2 & 4 a - 3 \\ 1 & 2 a - 2 & - 4 a + 3 & 6 a - b - 4 = 0 \end{array}$ $- 2 a + b = 4 . . . . (i)$ $6 a - b = 4 . . . . . (i i i)$ $(i) + (i i) :$ $4 a = 8 \Rightarrow a = 2$ $(i) \Rightarrow - 2 (2) + b = 4 \Rightarrow b = 8$
	Answered by mathzup last updated on 14/Jun/24
	$donc −1 est racine double de x^4 +2ax^3 −bx−5=u(x) ⇒u(−1)=0 et u^′ (−1)=0 ⇒ 1−2a+b−5=0 et 4(−1)^3 +6a(−1)^2 −b=0 ⇒ { ((−2a+b=4)),((−4+6a−b=0 ⇒ { ((b=4+2a)),((−4+6a−4−2a=0 ⇒)) :})) :} { ((b=2a+4)),((4a−8=0 ⇒ { ((a=2)),((b=8)) :})) :}$
	$d o n c - 1 e s t r a c i n e d o u b l e d e$ $x^{4} + 2 {a x}^{3} - b x - 5 = u (x)$ $\Rightarrow u (- 1) = 0 e t u^{^{'}} (- 1) = 0 \Rightarrow$ $1 - 2 a + b - 5 = 0 e t 4 {(- 1)}^{3} + 6 a {(- 1)}^{2} - b = 0 \Rightarrow$ ${\begin{cases} - 2 a + b = 4 \\ - 4 + 6 a - b = 0 \Rightarrow {\begin{cases} b = 4 + 2 a \\ - 4 + 6 a - 4 - 2 a = 0 \Rightarrow \end{cases} \end{cases}$ ${\begin{cases} b = 2 a + 4 \\ 4 a - 8 = 0 \Rightarrow {\begin{cases} a = 2 \\ b = 8 \end{cases} \end{cases}$
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