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Question Number 35602 by Rio Mike last updated on 20/May/18

$${\mathrm{Q}}_1.p\left(\mathrm{x}\right)={3\mathrm{x}}^3+{4\mathrm{x}}^2+5\mathrm{x}-\mathrm{k}\mathrm{and}$$$$(\mathrm{x}-1)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)\mathrm{find}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{and}\mathrm{the}\mathrm{remaining}$$$$\mathrm{two}\mathrm{factors}.$$$$Q_2.Evaluate\underset{r=1}{\sum ^{\mathrm{\infty }}}{3}^{2-r}.$$

Answered by Rasheed.Sindhi last updated on 21/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{If}\mathrm{p}\left(\mathrm{x}\right)\mathrm{is}\mathrm{divided}\mathrm{by}\mathrm{x}-\mathrm{r}\mathrm{then}\mathrm{the}$$$$\mathrm{remainder}\mathrm{is}\mathrm{p}\left(\mathrm{r}\right)\mathrm{and}\mathrm{if}\mathrm{x}-\mathrm{r}\mathrm{a}\mathrm{factor}$$$$\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)\mathrm{then}\mathrm{this}\mathrm{remainder}\mathrm{must}\mathrm{be}$$$$\mathrm{zero}.$$$$\mathrm{Here}\mathrm{x}-1\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)$$$$\therefore \mathrm{p}\left(1\right)=0$$$$\mathrm{Or}3{\left(1\right)}^3+4{\left(1\right)}^2+5\left(1\right)-\mathrm{k}=0$$$$\mathrm{k}=3+4+5=12$$$$\left(\begin{array}{ccccc}1)& 3& 4& 5& -12\\ & & 3& 7& 12\\ & 3& 7& 12& \overline{\mid 0}\end{array}\right)$$$$\mathrm{Quotient}={3\mathrm{x}}^2+7\mathrm{x}+12$$$$\mathrm{But}{3\mathrm{x}}^2+7\mathrm{x}+12\mathrm{is}\mathrm{not}\mathrm{factorable}$$$$\therefore \mathrm{Factors}:(\mathrm{x}-1)({3\mathrm{x}}^2+7\mathrm{x}+12)$$

Answered by Rasheed.Sindhi last updated on 21/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$3^{2-\mathrm{r}}\mathrm{is}\mathrm{a}\mathrm{GP}$$$$\mathrm{first}\mathrm{term}=\mathrm{a}=3^{2-1}=3$$$$\mathrm{common}\mathrm{ratio}=\frac{3^{2-2}}{3^{2-1}}=\frac{1}{3}$$$$\underset{r=1}{\sum ^{\mathrm{\infty }}}{3}^{2-r}=\frac{3}{1-\left(1/3\right)}=3\times \frac{3}{2}=\frac{9}{2}$$

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