MCQ
Expand the following binomials: $(x-3)^5$
 
  • A
    $x^5+25 x^4+90 x^3-270 x^2+405 x-243$
  • B
    $x^5-15 x^4+90 x^3-270 x^2-405 x-243$
  • C
    $x^5-15 x^4+80 x^3-270 x^2+405 x-243$
  • $x^5-15 x^4+90 x^3-270 x^2+405 x-243$

Answer

Correct option: D.
$x^5-15 x^4+90 x^3-270 x^2+405 x-243$
  1. $x^5-15 x^4+90 x^3-270 x^2+405 x-243$
Solution:
$(x-3)^5={ }^5 C_0 x^5+{ }^5 C_1 x^4(-3)^1+{ }^5 C_2 x^3(-3)^2+{ }^5 C_3 x^2(-3)^3+{ }^5 C_4 x(-3)^4+{ }^5 C_5(-3)^5$
$=x^5-15 x^4+90 x^3-270 x^2+405 x-243$

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