Plsss solve question 22 and 23..... urgent

Dear Student,
Solution:
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Question 22:
$$\begin{gathered} \mathrm{a}){\mathrm{x}}^3+9{\mathrm{x}}^2+23\mathrm{x}+15 \\ \mathrm{For} \mathrm{factorising} {\mathrm{x}}^3+9{\mathrm{x}}^2+23\mathrm{x}+15, \mathrm{we} \mathrm{should} \mathrm{know} \mathrm{atleast} \mathrm{one} \mathrm{zero} \mathrm{of} \mathrm{this} \mathrm{polynomial}. \mathrm{Once} \mathrm{we} \mathrm{know} \mathrm{it}, \mathrm{we} \mathrm{can} \mathrm{divide} \mathrm{the} \mathrm{polynomial} \mathrm{by} \mathrm{the} \mathrm{factor} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{quotient} \mathrm{and} \mathrm{factor} \mathrm{the} \mathrm{quotient} \mathrm{further} \mathrm{to} \mathrm{find} \mathrm{other} \mathrm{zeroes}. \\ \mathrm{Now}, \mathrm{Keeping} \mathrm{x}=-1, \\ \mathrm{p}(-1)={(-1)}^3+9{(-1)}^2+23(-1)+15 \\ \Rightarrow -1+9-23+15=24-24=0 \\ \mathrm{So}, (\mathrm{x}+1) \mathrm{is} \mathrm{a} \mathrm{factor}. \\ \mathrm{Now}, \mathrm{quotient} \mathrm{comes} \mathrm{out} \mathrm{to} \mathrm{be} = {\mathrm{x}}^{2 }+5\mathrm{x}+6 \\ \mathrm{Using} \mathrm{common} \mathrm{factor} \mathrm{theorem}, \\ {\mathrm{x}}^2+3\mathrm{x}+2\mathrm{x}+6 \\ \Rightarrow \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3) \\ \Rightarrow (\mathrm{x}+3\left)\right(\mathrm{x}+2) \\ \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{zeroes}, (\mathrm{x}+3\left)\right(\mathrm{x}+2)=0 \\ \mathrm{x}=-3,-2 \\ \mathrm{So}, \mathrm{the} \mathrm{zeroes} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{are} \mathrm{x}=-1,-2,-3 \\ \mathrm{Hence}, \mathrm{the} \mathrm{answer} \mathrm{should} \mathrm{be} (\mathrm{x}+1\left)\right(\mathrm{x}+2\left)\right(\mathrm{x}+3) \end{gathered}$$

Similarly, the second part can be attempted in the same manner.
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​​Question 23:
$$\begin{gathered} 23. 2\times 7^3-1\times 6^3-1\times 1^3 \\ \Rightarrow 2\times {(1+6)}^3-1\times 6^3-1\times 1^3 \\ \mathrm{Using} \mathrm{the} \mathrm{identity}: {(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3\mathrm{ab}(\mathrm{a}+\mathrm{b}) \\ \Rightarrow 2\times (1^3+6^3+3\times 1\times 6(1+6\left)\right)-1\times 6^3-1\times 1^3 \\ \Rightarrow 2\times 1^3+2\times 6^3+2\times 3\times 1\times 6\times 7-1\times 6^3-1\times 1^3 \\ \Rightarrow 1\times 1^{3 }+ 1\times 6^3 + 6\times 6\times 7 \\ \Rightarrow 1+216+252 \\ \Rightarrow 469 \end{gathered}$$
The solutions to part a of the 22nd question and 23rd question have been provided above.
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