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This is one of our many quizzes on Factoring monomials. Click on this to access more such competitive quizzes on Algebra.

Factoring monomials Quiz 2

mathematics

 Which type of equation will be given by the expression? $ \\ \dfrac{{(2k - 3)! - 2k!}}{{(2k - 3)!}}$

 $\dfrac{{(2k + 3)! - 2k!}}{{(2k)!}} = \dfrac{{(2k + 3)(2k + 2)(2k + 1)(2k)! - 2k!}}{{(2k)!}}  \\ \Rightarrow  \dfrac{{(2k)!((2k + 3)(2k + 2)(2k + 1) - 1)}}{{(2k)!}}  \\ \Rightarrow  \dfrac{{(2k)!}}{{(2k)!}}((2k + 3)(2k + 2)(2k + 1) - 1)  \\ \Rightarrow  ((2k + 3)(2k + 2)(2k + 1) - 1) \\  $ From the above equation it is cubic equation

 Calculate the number of roots and type of equation in the following expression. $ \\ \dfrac{1}{{2k!}}\left| {\begin{array}{*{20}{c}}  2&{(2k + 2)!} \\   3&{(2k + 1)!} \end{array}} \right| = 0$

 $\dfrac{1}{{2k!}}\left| {\begin{array}{*{20}{c}}  2&{(2k)!} \\   3&{(2k + 1)!} \end{array}} \right| = 0 \\  \dfrac{1}{{2k!}}(2 \times (2k + 1)!) - 3 \times (2k)!) = 0 \\  \dfrac{1}{{2k!}}(2(2k + 1)(2k)!) - 3 \times 2k!) = 0 \\  \dfrac{{2k!}}{{2k!}}(2(2k + 1) - 3) = 0 \\  2(2k + 1) = 3 \\  2k = 1.5 - 1 \\  k = 0.25 $

 Calculate the following expression $ \\ \mathop {\lim }\limits_{k \to 1} \dfrac{{((10k + 2)!) - ((10k + 1)!)}}{{(10k - 1)!}}$

 Use the property $k + n! = k + n - 1!$ and put the limits 

 $\mathop {\lim }\limits_{k \to 1} \dfrac{{((10k + 2)!) - ((10k + 1)!)}}{{(10k - 1)!}} \\  \mathop {\lim }\limits_{k \to 1} \dfrac{{((10k + 2)(10k + 1)!) - ((10k + 1)!)}}{{(10k - 1)!}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 1} (10k + 1)!\dfrac{{((10k + 2) - 1)}}{{(10k - 1)!}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 1} (10k + 1)(10k)(10k - 1)!\dfrac{{((10k + 2) - 1)}}{{(10k - 1)!}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 1} (10k + 1)(10k)((10k + 2) - 1) = (10 + 1)(10)((10 + 2) - 1)  \\ \Rightarrow  1210  $

 Expression is given below then the simplify value  $ \\ \dfrac{d}{{dk}}\dfrac{{({k^3} + 8)!}}{{({k^2} - 2k + 4)({k^2} + 6)!}}$

 $\dfrac{d}{{dk}}\dfrac{{({k^3} + 8)!}}{{({k^2} - 2k + 4)({k^3} + 6)!}} = \dfrac{d}{{dk}}\dfrac{{({k^3} + 8)({k^3} + 7)({k^3} + 6)!}}{{({k^2} - 2k + 4)({k^3} + 6)!}}  \\ \Rightarrow  \dfrac{d}{{dk}}\dfrac{{({k^3} + 8)({k^3} + 7)}}{{({k^2} - 2k + 4)}}   \\ \Rightarrow  \dfrac{d}{{dk}}\dfrac{{({k^3} + {2^3})({k^3} + 7)}}{{({k^2} - 2k + 4)}}  \\ \Rightarrow  \dfrac{d}{{dk}}\dfrac{{(k + 2)({k^2} - 2k + 4)({k^3} + 7)}}{{({k^2} - 2k + 4)}}  \\ \Rightarrow  \dfrac{d}{{dk}}(k + 2)({k^3} + 7)  \\ \Rightarrow  \dfrac{d}{{dk}}({k^4} + 7k + 2{k^3} + 14) = 4{k^3} + 6{k^2} + 7$

 Calculate the value of the equation ${k^2} + 5k + 6$, if the expression is following   $ \\ \mathop {\lim }\limits_{k \to 0} \dfrac{{(3k + 2){{((2k + 2)!)}^{12}}}}{{{{((2k - 1)!)}^{12}}{{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}}}$

 $\mathop {\lim }\limits_{k \to 0} \dfrac{{(3k + 2){{((2k + 2)!)}^{12}}}}{{{{((2k - 1)!)}^{12}}{{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 0} \dfrac{{(3k + 2)((2k + 2)(2k + 1)(2k)(2k - 1)!){)^{12}}}}{{{{((2k - 1)!)}^{12}}{{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 0} \dfrac{{(3k + 2){{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}(2k - 1)!){)^{12}}}}{{{{((2k - 1)!)}^{12}}{{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 0} \dfrac{{(3k + 2){{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}(2k - 1)!){)^{12}}}}{{{{((2k - 1)!)}^{12}}{{(2k + 2)}^{12}}{{(2k + 1)}^{12}}{{(2k)}^{12}}}}  \\ \Rightarrow  \mathop {\lim }\limits_{k \to 0} (3k + 2) = 2 $