MCQ
Expand the following binomials: (x - 3)5
    • A
      x5 + 25x4 + 90x3 - 270x2 + 405x - 243
    • B
      x5 - 15x4 + 90x3 - 270x2 - 405x - 243
    • C
      x5 - 15x4 + 80x3 - 270x2 + 405x - 243
    • D
      x5 - 15x4 + 90x3 - 270x2 + 405x - 243

    Answer

    1. x5 - 15x4 + 90x3 - 270x2 + 405x - 243

    Solution:

    (x - 3)5 = 5C0​x5 + 5C1​x(-3)1 + 5C2​x(-3)2 + 5C3​x(-3)3 + 5C4​x (-3)4 + 5C​(-3)5

    = x5 -15x4 + 90x3 - 270x2 + 405x - 243

    Need a full question paper?

    Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

    Start Generating Free

    Similar questions

    The position vectors of the four angular point of a tetrahedron OABC are (0, 0, 0), (0, 0, 2), (0, 4, 0) and (6, 0, 0) respectively. Find the coordinates of cenroid:
    If $\sqrt{3}\cos\text{x}+\sin\text{x}=\sqrt{2},$ then general value of $\theta$ is:
    1. $\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4},\ \text{n}\in\text{Z}$
    2. $(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
    3. $\text{n}\pi+\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
    4. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
    The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 8. The correct mean is:

    If in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15},\text{x}^{-17}$ occurs in rth term, then

    1. r = 10
    2. r = 11
    3. r = 12
    4. r = 13 
    If $z =(3+\sqrt{2} i)$ then $z \times z =$ ?
    If $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$ is a relation on Z, then the domain of R is:
    Let F1 be the set of all parallelograms, Fthe set of all rectangles, Fthe set of all rhombuses, F4 the set of all squares and Fthe set of trapeziums in a plane. Then F1 may be equal to:
    Find the mean of first six natural numbers:
    For any two sets A and B, $\text{A}\cap\text{(A}\cup\text{B)}'$ is equal to:

    Choose the correct answer.

    If the middle term of $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$ is equal to $7\frac{7}{8},$ then value of x is:

    1. $2\text{n}\pi+\frac{\pi}{6}.$

    2. $\text{n}\pi+\frac{\pi}{6}.$

    3. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$

    4. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}.$

    Hint: $\text{T}_6=\ ^{10}\text{C}_5\frac{1}{\text{x}^5}.\text{x}^5\ \sin^5\text{x}=\frac{63}{8}\Rightarrow\sin^5\text{x}=\frac{1}{2^5}\sin\frac{1}{2}$

    $\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$