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Fundamental Concepts Test - 6

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Fundamental Concepts Test - 6
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  • Question 1
    1 / -0

    Carry out the multiplication of expressions ineach of the following pairs. az, az

    Solution

    az×(a-z)

     = (az×a)+(az×(-z) )

    a2z2-az2

  • Question 2
    1 / -0

    Simplify: (1.5x-4a)(1.5x+4a+3)-4.5x+12

    Solution

    (1.5x-4a)(1.5x+4a+3)-4.5x+12a

    (1.5x-4a)(1.5x+4a+3)-3(1.5x-4a)

    (1.5x-4a)[(1.5x+4a)+(4a)+3-3]

    (1.5x-4a)(1.5x+4a)

    (1.5x)2-(4a)22.25x2-16a2.

  • Question 3
    1 / -0

    Find the product  (x+7a)(7xa)

    Solution

    x(7x-a)+7a(7xa)

    7x2-xa+49xa-7a2

    =7x2-7a2+48xa
  • Question 4
    1 / -0

    Simplify (m2+5)(b3+3)+5

    Solution

    multiplying both thebrackets with get m2b3+3m2+5b3+15+5

    m2b3+3m2+5b3+20
  • Question 5
    1 / -0

    Add: 2j2-4k2-l2,8j2+5k2+4j2,5j-5k2+2l2

    Solution

    say A=

    2j2-4k2-l2
    B=8j2+5k2+4l2
    C=5j2-5k2+2l2

    Now same variables with same exponents will be added so,

    =2j2+8j2+5j2=15j2,
    =-4k2+5k2+(-5k2)=-4k2,
    =-l2+4l2+2l2=5l2
    A+B+C=15j2-4k2+5l2

  • Question 6
    1 / -0

    Add all of them x8a,3xaax−8a,3xa−a and a+1

    Solution

    we have,
    x-8a,3xa-a and a+1
    On adding,
    (x-8a)+(3xa-a)+(a+1)

    x-8a+3xa-a+a+1
    x+3xa-8a+1

  • Question 7
    1 / -0

    What is the addition of the given variables :        p(pq),q(qr)p(p−q),q(q−r) & r(rp)

    Solution

    p(p-q)+q(q-r)+r(r-p)
    p2-pq+q2-qr+r2-pr
    =p2+q2+r2-pq-qr-rp

  • Question 8
    1 / -0
    Solve :-    xr+yr+xt+yt
    Solution

    xr+yr+xt+yt
    = r(x+y) + t(x+y)
    = (x+y) (r+t)

    Hence this is the correct answer.

  • Question 9
    1 / -0

    Divide 9h4by3h2

    Solution

    Let a = 9h4andb=3h2
    According to the question,

    ab=9h43h2

    ab=3h2×3h23h2

    ab=3h2

    Hence this is the answer.

  • Question 10
    1 / -0

    If a polynomial f(x) isdivided by c2-16 then the remainder is 5c+3, what will be the remainder when the same polynomial is divided by (c+4)?

    Solution

    Let the quotient be q(x)then,
    f(c)=(c2-16)+(5c+3)
    f(c)(c+4)=(c+4)(c-4)c+4q(c)+5c+3c+4
     Remainder when 5c+3 is divided by c+4 =5(-4)+3= -17

  • Question 11
    1 / -0

    Substract x+2b-c from 3x-b+2c

    Solution

    = 3x-b+2c-x-2b+c

    = 3x-b+2c-x-2b+c

    =3x-x-b-2b+2c+c

    2x-3b+3c.

  • Question 12
    1 / -0

    Find the product w2×(2w22)×(4w26)

    Solution

    All the terms have samebase, so we can add the powers 8w(2+22+26)=8w50

  • Question 13
    1 / -0

    Add 2d2+7d+5;3d+9;3d2-4d-3

    Solution

    we have,
    2d2+7d+5
    3d+9
    3d2-4d-3

    According to thequestion,
    =2d2+7d+5+3d+9+3d2-4d-3
    =5d2+6d+11

  • Question 14
    1 / -0

    Add (xa+az)2-2x2a2z Find the value when x=-1,a=1,z=2

    Solution

    (xa+az)2-2x2a2zwhen x= -1,a=1,z=2

    [a(x+z)]2-2x2a2z

    a2[x2+z2+2xz-2x2z]

    Now, putting the values inthe expression

    (1)2[(-1)2+22+2(-1)(2)-2(-1)2(2)]

    [1+4-4-4]=-3
  • Question 15
    1 / -0

    Multiply the binomial (f8)and (3f4)

    Solution

    f(3f-4)-8(3f-4)
    3f2-28f+32

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