• This is an assessment test.
  • To draw maximum benefit, study the concepts for the topic concerned.
  • Kindly take the tests in this series with a pre-defined schedule.

Algebra: Polynomials Test-4

Congratulations - you have completed Algebra: Polynomials Test-4.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1
$ \displaystyle If\,\,x+y=7,\,\,then\,\,the\,\,value\,\,\,of\,\,{{x}^{3}}+{{y}^{3}}+21xy\,\,\,is$
A
243
B
143
C
343
D
443
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}Given,\,\,x+y=7\\Now,\,\,{{x}^{3}}+{{y}^{3}}+21xy\\={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+21xy\\={{\left( 7 \right)}^{3}}-3xy\,\left( 7 \right)+21xy\\343-21xy+21xy=343\end{array}$
Question 2
$ \displaystyle If\,\,\,{{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}}={{z}^{\frac{1}{3}}},\,\,then\,\,\,\left[ {{\left( x+y-z \right)}^{3}}+27xyz \right]$
A
-1
B
1
C
0
D
27
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}{{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}}={{z}^{\frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( i \right)\\Cubing\,\,\,both\,\,\,sides,\,\\{{\left( {{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}} \right)}^{3}}=z\\\Rightarrow x+y+3\,\,\,{{x}^{\frac{1}{3}}}.y{{\,}^{\frac{1}{3}}}\,\,\left( {{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}} \right)=z\\=>x+y-z=3.{{x}^{\frac{1}{3}}}.{{y}^{\frac{1}{3}}}.{{z}^{\frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...............\left( ii \right)\\=>putting\text{ }in\text{ }equation\\{{\left( x+y-z \right)}^{3}}\,\,\,+27xyz=+27\,\,xyz-27xyz=0\end{array}$
Question 3
$ \displaystyle If\,\,\,x-\frac{1}{x}=4,\,\,\,then\,\,\left( x+\frac{1}{x} \right)$
A
$ \displaystyle 5\sqrt{2}$
B
$ \displaystyle 2\sqrt{5}$
C
$ \displaystyle 4\sqrt{2}$
D
$ \displaystyle 4\sqrt{5}$
Question 3 Explanation: 
$ \begin{array}{l}{{\left( x-\frac{1}{x} \right)}^{2}}=16~~~\\=>{{x}^{2}}+\frac{1}{{{x}^{2}}}=18+2=20\\=>x+\frac{1}{x}=2\sqrt{5}\end{array}$
Question 4
$ \displaystyle If\,\,\,x=3+\sqrt{8,\,}\,\,then\,\,the\,\,value\,\,of\,\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)$
A
34
B
24
C
38
D
36
Question 4 Explanation: 
$ \begin{array}{l}x+\frac{1}{x}=3+\sqrt{8}+\frac{1}{3+\sqrt{8}}=3+\sqrt{8}+\frac{3-\sqrt{8}}{\left( 3+\sqrt{8} \right)\left( 3-\sqrt{8} \right)}=6\\=>{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2\\=>36-2=34\end{array}$
Question 5
$ \displaystyle If\,\,\,4{{b}^{2}}+\frac{1}{{{b}^{2}}}=2,\,\,then\,\,\,the\,\,\,value\,\,of\,\,8{{b}^{3}}+\frac{1}{{{b}^{3}}}\,\,is$
A
0
B
1
C
2
D
5
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}{{\left( 2b+\frac{1}{b} \right)}^{2}}=4{{b}^{2}}+\frac{1}{{{b}^{2}}}+4=6\\\Rightarrow 2b+\frac{1}{b}=\sqrt{6}\\Therefore\,\,\,\,\,\\=>{{\left( 2b+\frac{1}{b} \right)}^{3}}=8{{b}^{3}}+\frac{1}{{{b}^{3}}}+3\times 2b\times \frac{1}{b}\left( 2b+\frac{1}{b} \right)\\=>\left( 6\sqrt{6} \right)=8{{b}^{3}}+\frac{1}{{{b}^{3}}}+6\sqrt{6}\\=>8{{b}^{3}}+\frac{1}{{{b}^{3}}}=0\end{array}$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect. Get Results
There are 5 questions to complete.
List
Return
Shaded items are complete.
12345
End
Return

Want to explore more Arithmetic Tests?

Explore Our Arithmetic Tests

Join our Free TELEGRAM GROUP for exclusive content and updates

Rsz 1rsz Close Img

Join Our Newsletter

Get the latest updates from our side, including offers and free live updates, on email.

Rsz Undraw Envelope N8lc Smal
Rsz 1rsz Close Img
Free Live Webinar Update