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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) = ((x^4 + 2ax^3 − bx − 5)/((x + 1)^2 )) Find: b = ?
$$\mathrm{P}\left(\mathrm{x}\right)\:\:\mathrm{is}\:\mathrm{polynomial} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{2ax}^{\mathrm{3}} \:−\:\mathrm{bx}\:−\:\mathrm{5}}{\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{b}}\:=\:? \\ $$
Answered by mr W last updated on 13/Jun/24
x^4 +2ax^3 −bx−5=(x+1)^2 (x^2 +px+q) x^4 +2ax^3 −bx−5=x^4 +(2+p)x^3 +(1+q+2p)x^2 +(p+2q)x+q q=−5 p+2q=−b ⇒b=8 ✓ 1+q+2p=0 ⇒p=2 2+p=2a ⇒a=2
$${x}^{\mathrm{4}} +\mathrm{2}{ax}^{\mathrm{3}} −{bx}−\mathrm{5}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{px}+{q}\right) \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{ax}^{\mathrm{3}} −{bx}−\mathrm{5}={x}^{\mathrm{4}} +\left(\mathrm{2}+{p}\right){x}^{\mathrm{3}} +\left(\mathrm{1}+{q}+\mathrm{2}{p}\right){x}^{\mathrm{2}} +\left({p}+\mathrm{2}{q}\right){x}+{q} \\ $$$${q}=−\mathrm{5} \\ $$$${p}+\mathrm{2}{q}=−{b}\:\Rightarrow{b}=\mathrm{8}\:\checkmark \\ $$$$\mathrm{1}+{q}+\mathrm{2}{p}=\mathrm{0}\:\Rightarrow{p}=\mathrm{2} \\ $$$$\mathrm{2}+{p}=\mathrm{2}{a}\:\Rightarrow{a}=\mathrm{2} \\ $$
Answered by efronzo1 last updated on 13/Jun/24
determinant ((,1,(2a),0,(−b),(−5)),((−1),∗,∗,(−1),(2−2a),(4a−3)),((−2),∗,(−2),(4−4a),(8a−6),∗),(,1,(2a−2),(3−4a),(6a−4−b),(4a−8))) ⇒4a−8=0 ⇒a=2 ⇒6(2)−4=b ⇒b=8
$$\:\:\:\:\:\:\:\begin{array}{|c|c|c|c|}{}&\hline{\mathrm{1}}&\hline{\mathrm{2a}}&\hline{\mathrm{0}}&\hline{−\mathrm{b}}&\hline{−\mathrm{5}}\\{−\mathrm{1}}&\hline{\ast}&\hline{\ast}&\hline{−\mathrm{1}}&\hline{\mathrm{2}−\mathrm{2a}}&\hline{\mathrm{4a}−\mathrm{3}}\\{−\mathrm{2}}&\hline{\ast}&\hline{−\mathrm{2}}&\hline{\mathrm{4}−\mathrm{4a}}&\hline{\mathrm{8a}−\mathrm{6}}&\hline{\ast}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{2a}−\mathrm{2}}&\hline{\mathrm{3}−\mathrm{4a}}&\hline{\mathrm{6a}−\mathrm{4}−\mathrm{b}}&\hline{\mathrm{4a}−\mathrm{8}}\\\hline\end{array} \\ $$$$\:\:\:\Rightarrow\mathrm{4a}−\mathrm{8}=\mathrm{0}\:\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\:\Rightarrow\mathrm{6}\left(\mathrm{2}\right)−\mathrm{4}=\mathrm{b}\:\Rightarrow\mathrm{b}=\mathrm{8}\: \\ $$
Answered by Rasheed.Sindhi last updated on 14/Jun/24
x^4 +2ax^3 −bx−5 is divisible by x+1 so by synthetic division: determinant (((−1)),1,(2a),0,(−b),(−5)),(,,(−1),(−2a+1),(2a−1),(−2a+b+1)),(,1,(2a−1),(−2a+1),(2a−b−1),(−2a+b−4=0))) Q(x)=x^3 +(2a−1)x^2 +(−2a+1)x+(2a−b−1) The quotient is again divisible by x+1 so again by synthetic division: determinant (((−1)),1,(2a−1),(−2a+1),(2a−b−1)),(,,(−1),(−2a+2),(4a−3)),(,1,(2a−2),(−4a+3),(6a−b−4=0))) −2a+b=4....(i) 6a−b=4.....(iii) (i)+(ii): 4a=8⇒a=2 (i)⇒−2(2)+b=4⇒b=8
$${x}^{\mathrm{4}} +\mathrm{2}{ax}^{\mathrm{3}} −{bx}−\mathrm{5}\:{is}\:{divisible}\:{by}\:{x}+\mathrm{1} \\ $$$${so}\:{by}\:{synthetic}\:{division}: \\ $$$$\begin{array}{|c|c|c|}{\left.−\mathrm{1}\right)}&\hline{\mathrm{1}}&\hline{\mathrm{2}{a}}&\hline{\mathrm{0}}&\hline{−{b}}&\hline{−\mathrm{5}}\\{}&\hline{}&\hline{−\mathrm{1}}&\hline{−\mathrm{2}{a}+\mathrm{1}}&\hline{\mathrm{2}{a}−\mathrm{1}}&\hline{−\mathrm{2}{a}+{b}+\mathrm{1}}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{2}{a}−\mathrm{1}}&\hline{−\mathrm{2}{a}+\mathrm{1}}&\hline{\mathrm{2}{a}−{b}−\mathrm{1}}&\hline{−\mathrm{2}{a}+{b}−\mathrm{4}=\mathrm{0}}\\\hline\end{array}\: \\ $$$${Q}\left({x}\right)={x}^{\mathrm{3}} +\left(\mathrm{2}{a}−\mathrm{1}\right){x}^{\mathrm{2}} +\left(−\mathrm{2}{a}+\mathrm{1}\right){x}+\left(\mathrm{2}{a}−{b}−\mathrm{1}\right) \\ $$$${The}\:{quotient}\:{is}\:{again}\:{divisible}\:{by}\:{x}+\mathrm{1} \\ $$$${so}\:{again}\:{by}\:{synthetic}\:{division}: \\ $$$$\begin{array}{|c|c|c|}{\left.−\mathrm{1}\right)}&\hline{\mathrm{1}}&\hline{\mathrm{2}{a}−\mathrm{1}}&\hline{−\mathrm{2}{a}+\mathrm{1}}&\hline{\mathrm{2}{a}−{b}−\mathrm{1}}\\{}&\hline{}&\hline{−\mathrm{1}}&\hline{−\mathrm{2}{a}+\mathrm{2}}&\hline{\mathrm{4}{a}−\mathrm{3}}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{2}{a}−\mathrm{2}}&\hline{−\mathrm{4}{a}+\mathrm{3}}&\hline{\mathrm{6}{a}−{b}−\mathrm{4}=\mathrm{0}}\\\hline\end{array}\: \\ $$$$−\mathrm{2}{a}+{b}=\mathrm{4}….\left({i}\right) \\ $$$$\:\:\:\mathrm{6}{a}−{b}=\mathrm{4}…..\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{4}{a}=\mathrm{8}\Rightarrow{a}=\mathrm{2} \\ $$$$\left({i}\right)\Rightarrow−\mathrm{2}\left(\mathrm{2}\right)+{b}=\mathrm{4}\Rightarrow{b}=\mathrm{8} \\ $$
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)) :})) :}
$${donc}\:−\mathrm{1}\:{est}\:{racine}\:{double}\:{de} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{ax}^{\mathrm{3}} −{bx}−\mathrm{5}={u}\left({x}\right) \\ $$$$\Rightarrow{u}\left(−\mathrm{1}\right)=\mathrm{0}\:{et}\:{u}^{'} \left(−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{1}−\mathrm{2}{a}+{b}−\mathrm{5}=\mathrm{0}\:{et}\:\mathrm{4}\left(−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{6}{a}\left(−\mathrm{1}\right)^{\mathrm{2}} −{b}=\mathrm{0}\:\Rightarrow \\ $$$$\begin{cases}{−\mathrm{2}{a}+{b}=\mathrm{4}}\\{−\mathrm{4}+\mathrm{6}{a}−{b}=\mathrm{0}\:\:\:\:\Rightarrow\begin{cases}{{b}=\mathrm{4}+\mathrm{2}{a}}\\{−\mathrm{4}+\mathrm{6}{a}−\mathrm{4}−\mathrm{2}{a}=\mathrm{0}\:\Rightarrow}\end{cases}}\end{cases} \\ $$$$\begin{cases}{{b}=\mathrm{2}{a}+\mathrm{4}}\\{\mathrm{4}{a}−\mathrm{8}=\mathrm{0}\:\:\Rightarrow\begin{cases}{{a}=\mathrm{2}}\\{{b}=\mathrm{8}}\end{cases}}\end{cases} \\ $$

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