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Q-1-p-x-3x-3-4x-2-5x-k-and-x-1-is-a-factor-of-p-x-find-the-value-of-k-and-the-remaining-two-factors-Q-2-Evaluate-r-1-3-2-r-

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Question Number 35602 by Rio Mike last updated on 20/May/18
Q_1 .p(x) = 3x^3 + 4x^2 +5x − k and (x−1) is a factor of p(x) find the value of k and the remaining two factors. Q_2 . Evaluate Σ_(r=1) ^∞ 3^(2−r) .
$$\mathrm{Q}_{\mathrm{1}} .{p}\left(\mathrm{x}\right)\:=\:\mathrm{3x}^{\mathrm{3}} +\:\mathrm{4x}^{\mathrm{2}} +\mathrm{5x}\:−\:\mathrm{k}\:\mathrm{and}\: \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining}\: \\ $$$$\mathrm{two}\:\mathrm{factors}. \\ $$$${Q}_{\mathrm{2}} .\:{Evaluate}\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}^{\mathrm{2}−{r}} . \\ $$
Answered by Rasheed.Sindhi last updated on 21/May/18
Q#1 If p(x) is divided by x−r then the remainder is p(r) and if x−r a factor of p(x) then this remainder must be zero. Here x−1 a factor of p(x) ∴ p(1)=0 Or 3(1)^3 + 4(1)^2 +5(1) − k=0 k=3+4+5=12 (((1)),3,4,5,(−12)),(,,3,7,( 12)),(,3,7,(12),( ∣ 0 ^(−) )) ) Quotient= 3x^2 +7x+12 But 3x^2 +7x+12 is not factorable ∴ Factors: (x−1)(3x^2 +7x+12)
$$\mathrm{Q}#\mathrm{1} \\ $$$$\mathrm{If}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}−\mathrm{r}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{remainder}\:\mathrm{is}\:\mathrm{p}\left(\mathrm{r}\right)\:\mathrm{and}\:\mathrm{if}\:\mathrm{x}−\mathrm{r}\:\mathrm{a}\:\mathrm{factor} \\ $$$$\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{then}\:\mathrm{this}\:\mathrm{remainder}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{zero}. \\ $$$$\:\mathrm{Here}\:\mathrm{x}−\mathrm{1}\:\mathrm{a}\:\mathrm{factor}\:\:\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right) \\ $$$$\therefore\:\mathrm{p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{Or}\:\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{3}} +\:\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{1}\right)\:−\:\mathrm{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{k}=\mathrm{3}+\mathrm{4}+\mathrm{5}=\mathrm{12} \\ $$$$\begin{pmatrix}{\left.\mathrm{1}\right)}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{−\mathrm{12}}\\{}&{}&{\mathrm{3}}&{\mathrm{7}}&{\:\:\:\:\mathrm{12}}\\{}&{\mathrm{3}}&{\mathrm{7}}&{\mathrm{12}}&{\:\:\:\:\:\overline {\mid\:\mathrm{0}\:\:}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Quotient}=\:\mathrm{3x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12} \\ $$$$\mathrm{But}\:\mathrm{3x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12}\:\mathrm{is}\:\mathrm{not}\:\mathrm{factorable} \\ $$$$\therefore\:\mathrm{Factors}:\:\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 21/May/18
Q#2 3^(2−r) is a GP first term =a=3^(2−1) =3 common ratio=(3^(2−2) /3^(2−1) )=(1/3) Σ_(r=1) ^∞ 3^(2−r) =(3/(1−(1/3)))=3×(3/2)=(9/2)
$$\mathrm{Q}#\mathrm{2} \\ $$$$\:\:\:\:\mathrm{3}^{\mathrm{2}−\mathrm{r}} \:\:\mathrm{is}\:\mathrm{a}\:\mathrm{GP} \\ $$$$\mathrm{first}\:\mathrm{term}\:=\mathrm{a}=\mathrm{3}^{\mathrm{2}−\mathrm{1}} =\mathrm{3} \\ $$$$\mathrm{common}\:\mathrm{ratio}=\frac{\mathrm{3}^{\mathrm{2}−\mathrm{2}} }{\mathrm{3}^{\mathrm{2}−\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}^{\mathrm{2}−{r}} =\frac{\mathrm{3}}{\mathrm{1}−\left(\mathrm{1}/\mathrm{3}\right)}=\mathrm{3}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$

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