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Question Number 175061 by daus last updated on 17/Aug/22

Commented by Rasheed.Sindhi last updated on 17/Aug/22

(a) 150

$$\left(\mathrm{a}\right)\:\:\mathrm{150} \\ $$

Answered by cortano1 last updated on 18/Aug/22

q(x)=p(x)(x^5 −2x+2) (a)⇒ { ((p(2)−5=0⇒p(2)=5)),((q(2)=p(2)(32−4+2))),((q(2)=5(30)=150)) :} (b)p(1)−5=0⇒p(1)=5 1+a+b=5⇒a+b=4 p(x)−5=x^2 +ax+b−5 x^2 +ax+b−5=(x−1)(x−2)=x^2 −3x+2 a=−3 ; b=5+2=7

$$\:{q}\left({x}\right)={p}\left({x}\right)\left({x}^{\mathrm{5}} −\mathrm{2}{x}+\mathrm{2}\right) \\ $$$$\:\:\left({a}\right)\Rightarrow\begin{cases}{{p}\left(\mathrm{2}\right)−\mathrm{5}=\mathrm{0}\Rightarrow{p}\left(\mathrm{2}\right)=\mathrm{5}}\\{{q}\left(\mathrm{2}\right)={p}\left(\mathrm{2}\right)\left(\mathrm{32}−\mathrm{4}+\mathrm{2}\right)}\\{{q}\left(\mathrm{2}\right)=\mathrm{5}\left(\mathrm{30}\right)=\mathrm{150}}\end{cases} \\ $$$$\left({b}\right){p}\left(\mathrm{1}\right)−\mathrm{5}=\mathrm{0}\Rightarrow{p}\left(\mathrm{1}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\:\mathrm{1}+{a}+{b}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{4} \\ $$$$\:{p}\left({x}\right)−\mathrm{5}={x}^{\mathrm{2}} +{ax}+{b}−\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{ax}+{b}−\mathrm{5}=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2} \\ $$$$\:{a}=−\mathrm{3}\:;\:{b}=\mathrm{5}+\mathrm{2}=\mathrm{7} \\ $$$$ \\ $$

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