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Question Number 73530 by Rio Michael last updated on 13/Nov/19

given the 3^(rd)  degree  polynomial  P(x) = (2x −1)(x−3)Q(x) + 12x−8  given that (x−1) is a factor of P(x) and  P(0) = 10  find Q(x)

$${given}\:{the}\:\mathrm{3}^{{rd}} \:{degree}\:\:{polynomial} \\ $$$${P}\left({x}\right)\:=\:\left(\mathrm{2}{x}\:−\mathrm{1}\right)\left({x}−\mathrm{3}\right){Q}\left({x}\right)\:+\:\mathrm{12}{x}−\mathrm{8} \\ $$$${given}\:{that}\:\left({x}−\mathrm{1}\right)\:{is}\:{a}\:{factor}\:{of}\:{P}\left({x}\right)\:{and}\:\:{P}\left(\mathrm{0}\right)\:=\:\mathrm{10} \\ $$$${find}\:{Q}\left({x}\right) \\ $$

Answered by MJS last updated on 13/Nov/19

y=(2x−1)(x−3)(ax+b)+12x−8  x=0  10=(−1)(−3)b−8 ⇒ b=6  (2x−1)(x−3)(ax+6)+12x−8  (x−1) is a factor ⇒ y=0 for x=1  0=(1)(−2)(a+6)+12−8 ⇒ a=−4  Q(x)=(−4x+6)  please check as I′m in a hurry

$${y}=\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({ax}+{b}\right)+\mathrm{12}{x}−\mathrm{8} \\ $$$${x}=\mathrm{0} \\ $$$$\mathrm{10}=\left(−\mathrm{1}\right)\left(−\mathrm{3}\right){b}−\mathrm{8}\:\Rightarrow\:{b}=\mathrm{6} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({ax}+\mathrm{6}\right)+\mathrm{12}{x}−\mathrm{8} \\ $$$$\left({x}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\Rightarrow\:{y}=\mathrm{0}\:\mathrm{for}\:{x}=\mathrm{1} \\ $$$$\mathrm{0}=\left(\mathrm{1}\right)\left(−\mathrm{2}\right)\left({a}+\mathrm{6}\right)+\mathrm{12}−\mathrm{8}\:\Rightarrow\:{a}=−\mathrm{4} \\ $$$${Q}\left({x}\right)=\left(−\mathrm{4}{x}+\mathrm{6}\right) \\ $$$$\mathrm{please}\:\mathrm{check}\:\mathrm{as}\:\mathrm{I}'\mathrm{m}\:\mathrm{in}\:\mathrm{a}\:\mathrm{hurry} \\ $$

Commented by Rio Michael last updated on 13/Nov/19

okay sir thanks i′ll check

$${okay}\:{sir}\:{thanks}\:{i}'{ll}\:{check} \\ $$

Commented by Rio Michael last updated on 13/Nov/19

correct sir thanks

$${correct}\:{sir}\:{thanks} \\ $$

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